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Physics 2 Marks Questions

Motion in a Straight Line · MP Board (MPBSE) STD 11 Science (Madhya Pradesh - English Medium). 100 questions.

Questions · Page 2 of 2

2 Marks Questions

Question 512 Marks
The distance travelled by a body is proportional to the square of time. What type of motion this body has?
Answer
Here, $\text{x}^2\propto \text{t}^2$ or $\text{x}=\text{kt}^2$ where , k is constant of proportinality. Now, $\text{v}=\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{kt}^2)=2\text{kt}$ and, $\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}}{}\text{dt}(2\text{kt})=2\text{k}$ (constant) Thus, the body has uniform accelerated motion.
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Question 522 Marks
In which of the following examples of motion, can the body be considered approximately a point object: A monkey sitting on top of a man cycling smoothly on a circular track.
Answer
The size of a monkey is very small as compared to the size of a circular track. Therefore, the monkey can be considered as a point sized object on the track.
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Question 532 Marks
A body goes from A to B with a velocity of 40m/s and comes back from B to A with a velocity of 60m/s. What is the (i) average velocity during the whole journey and (ii) average speed during the whole journey?
Answer
Average velocity $=\frac{\text{Total displacement}}{\text{times}}=0$ Since, there is no net displacement Average speed $=\frac{\text{Total distance}}{\text{time}}$ $=\frac{2\text{AB}}{\frac{\text{AB}}{40}+\frac{\text{AB}}{60}}=48\text{m/s}$
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Question 542 Marks
It is a common observation that rain clouds can be at about a kilometre altitude above the ground. Estimate the time required to flatten the drop.
Answer
Key concept: This problem can be solved by kinematic equations of motion and Newton’s second law that $\text{F}_\text{ext}=\frac{\text{dp}}{\text{dt}}$ will be used, where dp is change in momentum over time dt. Time required to flatten the drop = Time taken by the drop to travel the distance equal to the diameter of the drop near the ground. $\text{t}=\frac{\text{d}}{\text{v}}=\frac{4\times10^{-3}}{100\sqrt{2}}=0.028\times10^{-3}\text{s}$ $=2.8\times10^{-5}\text{s}=30\text{ms}$
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Question 552 Marks
A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time t is given by v(t) = 2t(3 - t); 0 < t < 3 and v(t) = -(t - 3)(6 - t) for 3 < t < 6s in m/s. It repeats this cycle till it reaches the height of 20m. At what time is its average velocity maximum?
Answer
Given velocity $\text{v}(\text{t})=2\text{t}(3-\text{t})=6\text{t}-2\text{t}^2$ From Eq. (i) $\text{v}=6\text{t}-2\text{t}^2$ $\Rightarrow\frac{\text{ds}}{\text{dt}}=6\text{t}-2\text{t}^2$ $\Rightarrow\text{ds}=(6\text{t}-2\text{t}^2)\text{dt}$ where, s is displacement $\therefore$ Distance travelled in time interval 0 to 3s, $\text{s}=\int^3_0(6\text{t}-2\text{t}^2)\text{dt}$ $=\Big[\frac{6\text{t}^2}{2}-\frac{2\text{t}^3}{3}\Big]^3_0=\Big[3\text{t}^2-\frac{2}{3}\text{t}^3\Big]^3_0$ $=3\times9-\frac{2}{3}\times3\times3\times3$ $=27-18=9\text{m}$ $\text{Average velocity}=\frac{\text{Distance travelled}}{\text{Time}}$ $=\frac{9}{3}=3\text{m/s}$ Given, $\text{x}=6\text{t}-2\text{t}^2$ $\Rightarrow3=6\text{t}-2\text{t}^2\Rightarrow2\text{t}^2-6\text{t}-3=0$ $\Rightarrow\text{t}=\frac{6\pm\sqrt{6^2-4\times2\times3}}{2\times2}=\frac{6\pm\sqrt{36-24}}{4}$ $=\frac{6\pm\sqrt{12}}{4}=\frac{3\pm2\sqrt{3}}{2}$ Considering positive sign only $\text{t}=\frac{3+2\sqrt{3}}{2}=\frac{3+2\times1.732}{2}=\frac{9}{4}\text{s}$
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Question 562 Marks
What is common between two graphs shown below?
Answer
Both graphs (a) and (b) represent positive and constant (different) velocity.
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Question 572 Marks
Acceleration is called as rate of change of velocity. Suppose we call rate of change of acceleration SLAP, what is the unit of SLAP?
Answer
SLAP $=\frac{\text{Acceleration}}{\text{Time}}$ $=\frac{\text{ms}^{-2}}{\text{s}}=\text{ms}^{-3}$
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Question 582 Marks
A body travels with a velocity $\mathrm{v}_1$ for time $\mathrm{t}_1$ and with a velocity $\mathrm{v}_2$ for time $\mathrm{t}_2$, find the average velocity of the body for the total time.
Answer
Displacement travelled in time $(\text{t}_1+\text{t}_2)=\text{s}_1+\text{s}_2$ $=\text{v}_1\text{t}_1+\text{v}_2\text{t}_2$ $\therefore$ Average velocity $=\frac{\text{Net displacement}}{\text{Total time taken}}$ $=\frac{\text{v}_1\text{t}_1+\text{v}_2\text{t}_2}{(\text{t}_1+\text{t}_2)}$
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Question 592 Marks
The electric current in a discharging $R-C$ circuit is given by $\text{i}=\text{i}_0\text{e}^{-\frac{\text{t}}{\text{RC}}}$ where $i_0, R$ and $C$ are constant parameters and $t$ is time. Let $\text{i}_0=2.00\text{A},\text{R}=600\times10^5\Omega$ and $\text{C}=0.500\mu\text{F}.$
  1. Find the current at $t = 0.3s.$
  2. Find the rate of change of current at $t = 0.3s.$
  3. Find approximately the current at $t = 0.31s.$
Answer
Equation $\text{i = i}_0\text{e}^{-\frac{\text{t}}{\text{RC}}}$
$\text{i}_0=2\text{A, R}=6\times10^{-5}\Omega,\text{C}=0.0500\times10^{-6},\text{F}=5\times10^{-7}\text{F}$
  1. $\text{i}=2\times\text{e}^{\Big(\frac{-0.3}{6\times0^3\times5\times10^{-7}}\Big)}=2\times\text{e}^{\big(\frac{-0.3}{0.3}\big)}=\frac{2}{\text{e}}\text{amp}$
  2. $\frac{\text{di}}{\text{dt}}=\frac{-\text{i}_0}{\text{RC}}\text{e}^{-\frac{\text{t}}{\text{RC}}}$ when $t = 0.3$
$\sec\Rightarrow\frac{\text{di}}{\text{dt}}=-\frac{2}{0.30}\text{e}^{\big(\frac{-03}{0.3}\big)}=\frac{-20}{3\text{e}}\text{amp/sec}$
  1. At $\text{t}=0.31\text{ sec},\text{i}=2\text{e}^{\big(-\frac{-0.3}{0.3}\big)}=\frac{5.8}{3\text{e}}\text{amp}.$
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Question 602 Marks
A car travelling with a speed of 90km/ h on a straight road is ahead of a scooter travelling with a speed of 60km/ h. How would the relative velocity be altered, if scooter is ahead of the car?
Answer
Let $\mathrm{v}_{\mathrm{c}}$ and $\mathrm{v}_{\mathrm{s}}$ be the velocities of the car and the scooter, respectively. $\mathrm{v}_{\mathrm{C}}=90 \mathrm{~km} / \mathrm{h}$ and $\mathrm{v}_{\mathrm{s}}=60 \mathrm{~km} / \mathrm{h}$ (given) When the car is ahead of the scooter, then the relative velocity is $\mathrm{V}_{\mathrm{CS}}=\mathrm{v}_{\mathrm{C}}-\mathrm{v}_{\mathrm{s}}=90-60=30 \mathrm{~km} / \mathrm{h}$ (away from the scooter) When the scooter is ahead of the car, then the relative velocity is $\mathrm{V}_{\mathrm{Cs}}=\mathrm{v} .-\mathrm{v}_r=90-60=30 \mathrm{~km} / \mathrm{h}$ (towards the scooter)
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Question 612 Marks
A car travelling at a speed of 10m/s due North, turns to its left and travels with same speed. Find the change in velocity associated.
Answer
Change in velocity, $\text{v}_\text{f}-\text{v}_\text{i}=\vec{\text{v}}_\text{W}-\vec{\text{v}}_\text{N}=\vec{\text{v}}_\text{W}\\+(-\vec{\text{v}}_\text{N})=\vec{\text{v}}_\text{W}+\vec{\text{v}}_{\text{S}}$ Since magnitudes are equal the change in velocity $=\sqrt{2}\text{v}_\text{W}=10\sqrt{2}$ in south westerly direction.
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Question 622 Marks
The graph between total path length and time for a particle moving along a straight line as shown in figure is not possible. Explain why?
Answer
The graph shows that with the passage of time, total path length first increases and then decreases. The path length always increases or remains constant with passage of time and it does not decrease with time as shown in figure. Thus, this graph is not possible.
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Question 632 Marks
Which of the following is true for displacement?
  1. It cannot be zero.
  2. Its magnitude is greater than the distance travelled by the object.
Answer
Both these statements are not true, because:
  1. Its magnitude can be zero.
  2. Its magnitude is either less than or equal to the distance travelled by the object.
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Question 642 Marks
If a particle is accelerating, it is either speeding up or speeding down. Do you agree with this statement?
Answer
Acceleration doesn’t mean speeding up or down. It means change of velocity either change in magnitude or direction.
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Question 652 Marks
Find the area bounded under the curve $y = 3x^2 + 6x + 7$ and the X-axis with the ordinates at $x = 5$ and $x = 10$.
Answer
$y = 3x^2 + 6x + 7 \therefore$ Area bounded by the curve, x-axis with coordinates with $x = 5$ and $x = 10$ is given by, $\text{Area}=\int\limits^{\text{y}}_{0}\text{dy}=\int\limits^{10}_5(3\text{x}^2+6\text{x}+7)\text{dx}\\=3\frac{\text{x}^3}{3}\Big]^{10}_5+5\frac{\text{x}^2}{3}\Big]^{10}_5+7\text{x}]^{10}_{5}=1135\text{sq. units}$
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Question 662 Marks
An object is moving with uniform velocity $v$ along a straight line. What will be the shape of displacement$-$time graph if:
  1. $x_0 = -ve, v = +ve$,
  2. $x_0 = +ve, v = -ve$
Here, $x_0$ represents the position of the particle at time $t = 0$.
Answer
Displacement$-$time graphs are shown in Fig.
In Fig. $(a)\ x_0$ is $-ve$ but velocity $v$ is $+ve$
i.e., slope of curve is $+ve.$
In Fig. $(b), x_0$ is $+ve$ but slope of line and hence velocity is negative.
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Question 672 Marks
A body is travelling in a straight line with a uniformly increasing speed. Plot a graph which represents the change in distance (s) travelled with time (t).
Answer
Let a body is travelling with initial speed ‘u’ in a straight line, then using relation, $\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$ For u = 0; $\text{s}=\frac{1}{2}\text{at}^2$ $\text{s}\propto \text{t}^2$
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Question 682 Marks
What is meant by 'point object' in physics?
Answer
An object is said to be point object if its dimensions are negligible as compared to the distance travelled by it. For example, an aeroplane which flies from Delhi to London.
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Question 692 Marks
A man walks on a straight road from his home to a market 2.5 km away with a speed of $5 \mathrm{~km}{ }^{\mathrm{h}-1}$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5 \mathrm{~km} \mathrm{~h}^{-1}$. What is the Magnitude of average velocity, and.
Answer
Time taken by the man to reach the market from home $\mathrm{t}_1=2.5 / 5=1 / 2 \mathrm{~h}=30 \mathrm{~min}$ Time taken by the man to reach home from the market, $\mathrm{t}_2=2.5 / 7.5=1 / 3 \mathrm{~h}=20 \mathrm{~min}$ Total time taken in the whole journey $=30+20=50 \mathrm{~min}$
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Question 702 Marks
A particle is moving in a straight line. Is it possible for it to maintain the motion in the same direction while the acceleration is in the reverse direction?
Answer
Yes, due to acceleration in reverse direction the velocity starts decreasing with time but the direction of motion is maintained till the velocity is reduced to zero.
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Question 722 Marks
At t = 0, a particle is at rest at origin. Its acceleration is $2m/s^2$ for the first 3s and $-2m/s^2$ for next 3s. Plot the acceleration versus time and velocity versus time graph.
Answer
The acceleration-time graph is The area enclosed between a-t curve gives change in velocity for the corresponding interval. At t = 0, v = 0, hence final velocity at t = 3s will increase to 6m/s. In next 3s, the velocity will decrease to zero. Thus, the velocity-time graph is
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Question 732 Marks
A ball is thrown vertically up with a velocity of 20m/s. Construct acceleration time and displacement time graph.
Answer
$\mathrm{u}=20 \mathrm{~ms}^{-1}, \mathrm{a}=\mathrm{gms}^{-2}$ Time to reach the highest point, $\text{t}=\frac{\text{u}}{\text{g}}=2\text{ seconds}.$ Max. height $=\frac{1}{2}\text{g}\times4=20\text{m}$
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Question 742 Marks
In which of the following examples of motion, can the body be considered approximately a point object: A railway carriage moving without jerks between two stations.
Answer
The size of a carriage is very small as compared to the distance between two stations. Therefore, the carriage can be treated as a point sized object.
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Question 752 Marks
The position$-$time$-$graph in figure depicts the journey of three bodies $A, B$ and $C.$
  1. At $1s$, which has the greatest velocity?
  2. At $2s$, which has travelled the farthest?
  3. When $A$ meets $C,$ is $B$ moving faster or slower than $A?$
  4. Is there any time at which the velocity of $A$ is equal to that of $B?$
Answer
  1. $B$
  2. $C$
  3. Slower
  4. Yes, in the interval $2$ to $3s$
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Question 762 Marks
The velocity of a particle is given by equation $v=4+2\left(c_1+c_2 t\right)$, where $c_1$ and $c_2$ care constant. Find the initial velocity and acceleration of the particle.
Answer
Given equation of velocity, $v=4+2\left(c_1+c_2 t\right) \Rightarrow v=\left(4+2 c_1\right)+2 c_2 t$ Compare the above equation with equation of motion $\mathrm{v}=\mathrm{u}+$ at Initial velocity, $\mathrm{u}=4+2 \mathrm{c}_1$ Acceleration of the Particle $=2 \mathrm{c}_2$.
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Question 772 Marks
A body moving with a uniform acceleration describes $12m$ in $3^{rd}$ second of its motion and $20m$ in the $5^{th}$ second. Find the velocity after $10$ seconds.
Answer
Let the initial velocity of the body 'u' and acceleration 'a'. Using equation for $n^{th}$ second motion $\text{S}_{\text{n}^{\text{th}}}=\text{u}+\frac{1}{2}\text{a}(2\text{n}-1)$
$\text{S}_{3^\text{rd}}=\text{u}+\frac{1}{2}\text{a}(2\times3-1)$
$12=\text{u}+\frac{5}{2}\text{a}\ \dots(\text{i})$
$\text{S}_{5^\text{th}}=\text{u}+\frac{1}{2}\text{a}(2\times5-1)$
$20=\text{u}+\frac{9}{2}\text{a}\ \dots(\text{ii})$ By solving equations (i) and (ii), we get $a = 4ms^{-2}$ and $u = 2ms^{-1}$ Now, Using equation $v = u +$ at v after 10 second = $2 + 4 \times 10ms v = 42ms^{-1}$.
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Question 782 Marks
Two particles A and B are moving along the same straight line. B is ahead of A. Velocities remaining unchanged, what would be the effect on the magnitude of relative velocity if A is ahead of B?
Answer
There will be no effect on the magnitude of relative velocity because relative velocity is $(\vec{\text{v}}_{\text{A}}-\vec{\text{v}}_{\text{B}})$ which always remains constant.
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Question 792 Marks
Two bodies of different masses m, and m, are dropped from two different heights 'a' and 'b'. What is the ratio of time taken by the two to drop through these distances?
Answer
$\text{a}=\frac{1}{2}\text{gt}^2_1$ $\text{t}_1=\sqrt{\frac{2\text{a}}{\text{g}}}$ $\text{b}=\frac{1}{2}\text{gt}^2_2$ $\text{t}_2=\sqrt{\frac{2\text{b}}{\text{g}}}$ $\therefore \frac{\text{t}_1}{\text{t}_2}=\frac{\sqrt{\text{a}}}{\sqrt{\text{b}}}$
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Question 802 Marks
The driver of a truck travelling with a velocity v suddenly notices a brick wall in front of him at a distance d. Is it better for him to apply brakes or to make a circular turn without applying brakes in order to just avoid crashing into the wall? Why?
Answer
$\text{F}_\text{B}\times\text{d}=\frac{1}{2}\text{mv}^2$ $\text{F}_\text{B}=\frac{\text{mv}^2}{2\text{d}}$ $\text{F}_\text{T}=\frac{\text{mv}^2}{\text{d}}$ $\therefore \text{F}_\text{T}=2\text{F}_\text{B}$
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Question 812 Marks
It is a common observation that rain clouds can be at about a kilometre altitude above the ground.A typical rain drop is about 4mm diameter. Momentum is mass x speed in magnitude. Estimate its momentum when it hits ground.
Answer
Key concept: This problem can be solved by kinematic equations of motion and Newton's second law that $\mathrm{F}_{\mathrm{ext}}=\frac{\mathrm{dp}}{\mathrm{dt}}$ will be used, where dp is change in momentum over time dt . Diameter of the $\mathrm{drop}(\mathrm{d})=2 \mathrm{r}=4 \mathrm{~mm}$ Radius of the drop $(\mathrm{r})=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$ Mass of a rain $\operatorname{drop}(\mathrm{m})=\mathrm{v} \times \mathrm{p}$ $=\frac{4}{3} \pi \mathrm{r}^3 \mathrm{p}=\frac{4}{3} \times \frac{22}{7} \times\left(2 \times 10^{-3}\right)^3 \times 10^3\left(\because\right.$ Density of water $\left.=10^3 \mathrm{~kg} / \mathrm{m}^2\right) \Rightarrow \mathrm{m}=3.4 \times 10^{-5} \mathrm{~kg}$ Momentum of the rain $\operatorname{drop}(\mathrm{p})=\mathrm{mv}=3.4 \times 10^{-5} \times 100 \sqrt{2} \Rightarrow \mathrm{p}=4.7 \times 10^{-3} \mathrm{~kg}-\mathrm{m} / \mathrm{s}=5 \times 10^{-3} \mathrm{~kg}-\mathrm{m} / \mathrm{s}$
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Question 822 Marks
If position of a particle at instant t is given by $\text{x}=\text{t}^3$, find acceleration of the particle.
Answer
Given, $\text{x}=\text{t}^3$ $\therefore \nu=\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{t}^3)=3\text{t}^2$ Now, acceleration (a) $=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(3\text{t}^2)=6\text{t}.$
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Question 832 Marks
Write two important points to distinguish displacement from distance.
Answer
Length of actual path covered between the initial and final points is distance while the length of the shortest path between initial and final points is displacement. The magnitude of displacement can be both positive and negative while distance is always positive.
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Question 842 Marks
The velocity of a particle is given by equation $v=4+2\left(C_1+C_2 t\right)$ where, $C$, and $C$, are constant. Find the initial velocity and acceleration of the particle.
Answer
The given equation is $v=4+2\left(C_1+C_2 t\right) \Rightarrow v=\left(4+2 C_1\right)+2 C_2 t$ Comparing the above equations with equation of motion $\mathrm{v}=\mathrm{u}+$ at Initial velocity, $\mathrm{u}=4+2 \mathrm{C}_1$ Acceleration of the particle $=2 \mathrm{C}_2$
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Question 852 Marks
A body starts from rest and moves along a straight line. It has uniformly accelerated motion upto time $t_1$. During the interval $t_2-t_1$ it moves with uniform velocity. After time $t_2$ its motion is retarded, and it comes to rest at time $t_3$. Draw the velocity-time graph.
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Question 862 Marks
The odometer of Raja's car reads 1700km at the start of a trip and 2500km at the end of the trip. The trip took 16h. What is the average speed of Raja's car in $ms^{-1}$?
Answer
Here, distance covered by the car = (2500 -1700)km = 800km Time elapsed = 16h Average speed $=\frac{\text{Total distance}}{\text{Total time}}$ $\text{u}=\frac{800}{16}=50\text{km/h}$ As 1km = 1000m and 1h = 3600s $\text{u}=\frac{50\times1000}{1\times3600}=13.9\text{ms}^{-1}$
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Question 872 Marks
The velocity of a particle is $v=5+2\left(a_1+a_2 t\right)$ where $a_1$ and $a_2$ are constants and $t$ is the time. What is the acceleration of the particle?
Answer
$\text{v}=5+2(\text{a}_1+\text{a}_2\text{t})$ $\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}}{\text{dt}}[5+2(\text{a}_1+\text{a}_2\text{t})]$ $\text{a}=2\text{a}_2$
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Question 882 Marks
A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time t is given by v(t) = 2t(3 - t); 0 < t < 3 and v(t) = -(t - 3)(6 - t) for 3 < t < 6s in m/s. It repeats this cycle till it reaches the height of 20m. At what time is its velocity maximum?
Answer
Given velocity $\text{v}(\text{t})=2\text{t}(3-\text{t})=6\text{t}-2\text{t}^2$ For maximum velocity $\frac{\text{dv(t)}}{\text{dt}}=0$ $\Rightarrow\frac{\text{d}}{\text{dt}}(6\text{t}-2\text{t}^2)=0$ $\Rightarrow6-4\text{t}=0$ $\Rightarrow\text{t}=\frac{6}{4}=\frac{3}{2}\text{s}=1.5\text{s}$
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Question 892 Marks
Points P, Q and R are in a vertical line such that PQ = QR. A ball at P is allowed to fall freely. What is the ratio of the times of descent through PQ and QR?
Answer
Let t, and tą be the times of descent through PQ and QR, respectively. Let PQ = QR = h Then, $\text{h}=\frac{1}{2}\text{gt}^2_1$ and $2\text{h}=\frac{1}{2}\text{g}(\text{t}_1+\text{t}_2)^2$ By dividing, we get $\frac{1}{2}=\frac{\text{t}^2_1}{(\text{t}_1+\text{t}_2)^2}$ $\frac{1}{\sqrt{2}}=\frac{\text{t}_1}{\text{t}_1+\text{t}_2}$ Hence, $\text{t}_1:\text{t}_2=2:(\sqrt{2}-1)$
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Question 902 Marks
Find the area bounded by the curve $y = e^{-x}$, the X-axis and the Y-axis.
Answer
The given function is $y = e^{-x}$ When $x = 0, y = e^{-0} = 1$ x increases, y value deceases and only at $\text{x}=\infty,\text{y} = 0.$ So, the required area can be found out by integrating the function from 0 to $\infty.$ So, $\text{Area}=\int\limits^{\infty}_0\text{e}^{\text{-x}}\text{dx}=-[\text{e}^{-\text{x}}]^{\infty}_0=1.$
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Question 912 Marks
Two trains of lengths 109 m and 91 m are moving in opposite directions with velocities $34 \mathrm{~km} \mathrm{~h}^{-1}$ and $38 \mathrm{~km} / \mathrm{h}^{-1}$ respectively. In what time the two trains will completely cross each other? Choose the most logical reference point for time measurement.
Answer
Relative speed $=(34+38) \mathrm{km} / \mathrm{h}^{-1}=72 \mathrm{~km} / \mathrm{h}^{-1}=72 \times \frac{5}{18} \mathrm{~ms}^{-1}=20 \mathrm{~ms}^{-1}$ Total distance $=(109+91) \mathrm{m}=200 \mathrm{~m}$ Time $=\frac{200 \mathrm{~m}}{20 \mathrm{~ms}^{-1}}=10 \mathrm{~s}$
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Question 922 Marks
Explain the difference between uniform velocity and variable velocity.
Answer
If a body travels equal displacements in equal intervals of time, then the velocity of body is uniform velocity. On the other hand, if the body covers unequal displacements in equal intervals of time, then its velocity is variable velocity.
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Question 932 Marks
Is earth inertial or non-inertial frame of reference?
Answer
Since, earth revolves around the sun and also spins about its own axis, so it is an accelerated frame of reference. Hence, earth is a non-inertial frame of reference. However, if we do not take large scale motion such as wind and ocean currents into consideration, we can say that approximation the earth is an inertial frame.
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Question 942 Marks
Is the time variation of position shown in figure observed in nature possible?
Answer
No, when x increases, the time first increases and then decreases. It is not possible. It implies that at a given time, the body in motion is simultaneously at two different positions which is not possible.
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Question 952 Marks
A particle starts from rest, and its acceleration (a) plotted against time (t) is shown here. Plot the corresponding velocity (v) against time (t). Also plot the corresponding displacement (s) against time (t).
Answer
a = constant $\text{v}\propto \text{t}$ $\text{s}\propto \text{t}^2$ The corresponding graphs are:
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Question 962 Marks
The data regarding the motion of two different objects Pand Q are given in the following table. Examine them carefully and state whether the motion of the objects is uniform or non-uniform.
Time
Distance travelled by object P (in m)
Distance travelled by object Q (in m)
9:30 am
10
12
9:45 am
20
19
10:00 am
30
23
10:15 am
40
35
10:30 am
50
37
10.45 am
60
41
11:00 am
70
44
Answer
We can see that the object P covers a distance of 10m in every 15 min. In other words, it covers equal distance in equal intervals of time. So, the motion of object Pis uniform. On the other hand, the object Q covers 7 m from 9:30 am to 9:45 am, 4 m from 9:45 am to 10:00 am and so on. In other words, it covers unequal distances in equal intervals of time. So, the motion of object Q is non-uniform.
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Question 972 Marks
Explain how an object could have zero average velocity but non-zero average speed?
Answer
Average velocity, $\text{v}_{\text{av}}=\frac{\text{Net displacement}}{\text{Total time taken}}$ and average speed, $\text{s}_{\text{av}}=\frac{\text{Total distanve travelled}}{\text{Total time taken}}$ If an object moves along a straight line starting from origin and then returns back to origin. Average velocity = 0 And Average Speed $=\frac{2\text{s}}{\text{t}}$
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Question 982 Marks
The distance covered by a body is found to be directly proportional to the square of time. Is the body moving with uniform velocity or uniform acceleration? If the distance travelled be directly proportional to time.
Answer
Let $\text{s}\propto \text{t}^2$ $\text{s}=\text{kt}^2$ $\text{v}=\frac{\text{ds}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{kt})^2=2\text{kt},$ where k is costant and $\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{v})$ $=\frac{\text{d}}{\text{dt}}(2\text{kt})=2\text{k}$ $\therefore$ The body is moving with uniform acceleration. In case $\text{s}\propto \text{t, s}= \text{kt}$ or $\text{v}=\frac{\text{ds}}{\text{dt}}=\text{k}$ (constant) and the body is moving with uniform velocity.
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Question 992 Marks
What are uses of a velocity$-$time graph?
Answer
From a velocity$-$time graph, we can find out:
  1. The velocity of a body at any instant.
  2. The acceleration of the body and
  3. The net displacement of the body in a given time-interval.
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Question 1002 Marks
The position of an object moving along $x$-axis is given by $x=a+b t^2$ where $a=8.5 m , b=2.5 m s ^{-2}$ and $t$ is measured in seconds. What is its velocity at $t=0 s$ and $t=2.0 s$. What is the average velocity between $t=2.0 s$ and $t=4.0 s$ ?
Answer
In notation of differential calculus, the velocity is
$
v=\frac{d x}{d t}=\frac{d}{d t}\left(a+b t^2\right)=2 b t=5.0 t m s ^{-1}
$
At $t=0 s , \quad V=0 m s ^{-1}$ and at $t=2.0 s$, $V=10 m s ^{-1}$.
Average velocity $=\frac{x(4.0)-x(2.0)}{4.0-2.0}$
$\begin{aligned}=\frac{a+16 b-a-4 b}{2.0} & =6.0 \times b \\ =6.0 \times 2.5 & =15 m s ^{-1}\end{aligned}$
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