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Science Que-Ans (Each of 2 Mark )

Motion · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 87 questions.

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Que-Ans (Each of 2 Mark )

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Question 12 Marks
A stone is thrown in a vertically upward direction with a velocity of 5ms-1. If the acceleration of the stone during its motion is 10ms–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answer
Initial velocity of stone = u = 5ms-1
Final velocity of stone = v = 0ms-1 (since at the maximum height, the stone comes to rest)
Acceleration = a = 10ms.-2
Using the first equation of motion,
v = u + at
0 = 5 + (-10)t (because the acceleration is in the opposite direction to the motion)
t = 0.5 sec
Using this answer in the second equation of motion,
s = ut + 0.5at2
s = (5 * 0.5) + (0.5 * (-10) * 0.52)
s = (2.5) – (1.25)
s = 1.25 metres
Thus, the stone attains 1.25m at the maximum accelerating point.
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Question 22 Marks
A racing car has a uniform acceleration of 4ms-2. What distance will it cover in 10s after start?
Answer
Initial velocity = u = 0ms-1
Acceleration = a = 4ms-2
Time = t = 10 sec
Using the second equation of motion,
S = ut + 0.5 at2
S = 0 = (0.5 * 4 * 102)
S = 200m
Hence the horse covers 200 metres in the first 10 seconds.
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Question 32 Marks
A trolley, while going down an inclined plane, has an acceleration of 2cm s-2. What will be its velocity 3 s after the start?
Answer
Initial velocity = 0m/ s (since the trolley was at rest)
Acceleration = a = 2cm s-2 = 0.02ms-2
Time = t = 3 sec
Using the first equation of motion,
v = u + at
v = 0 + (0.02 * 3)
v = 0.06ms-2
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Question 42 Marks
A train is travelling at a speed of 90km/h–1. Brakes are applied so as to produce a uniform acceleration of – 0.5ms-2. Find how far the train will go before it is brought to rest.
Answer
Initial speed of the car: u = 90km/h-1 = 25ms-1
Final speed of the car: v = 0ms-1 (since it comes to rest)
Acceleration: -0.5 ms-2
Using the third equation of motion
v2 = u2 + 2as
(0)2 = (25)2 + 2(-0.5)s
On solving the above equation,
S = 625m
Therefore, the car stops (1000 – 625 ) = 375m before the block.
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Question 52 Marks
A bus starting from rest moves with a uniform acceleration of 0.1m/ s-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Answer
  1. Given:

u = 0

v = ?

a = 0.1 m/s2

t = 120 sec

a = v - ut

$0.1 = \frac{\text{v}–0}{120}$

= 12m/ s

  1. Given:

u = 0

v = 12m/ s

t = 120 sec

a = 0.1m/ s2

According to the third equation of motion,

v2 + u2 = 2 as

(12)2 + (0)2 = 2(0.1)s

On solving the above equation,

s = 720 metres.

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Question 62 Marks
What can you say about the motion of an object if its speed - time graph is a straight line parallel to the time axis?
Answer
If speed time graph is a straight line parallel to the time axis, the object is moving uniformly.

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Question 72 Marks
What can you say about the motion of an object whose distance - time graph is a straight line parallel to the time axis?
Answer
If distance time graph is a straight line parallel to the time axis, the body is at rest.

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Question 82 Marks
What is the nature of the distance - time graphs for uniform and non-uniform motion of an object?
Answer
When the motion is uniform,the distance time graph is a straight line with a slope.

When the motion is non uniform, the distance time graph is not a straight line.It can be any curve.

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Question 92 Marks
A train starting from a railway station and moving with uniform acceleration attains a speed 40km/h−1 in 10 minutes. Find its acceleration.
Answer
Initial velocity of the train, u = 0 (since the train is initially at rest)

Final velocity of the train, v = 40km/h

$=40\times\frac{5}{18}=11.11\text{m/s}$

Time taken, t = 10 min = 10 × 60 = 600s

Acceleration, $\text{a}=\frac{\text{v}-\text{u}}{\text{t}}=\frac{11.11-0}{600}$

$0.0185\text{m/s}^2$

Here, the negative sign of acceleration indicates that the velocity of the car is decreasing.

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Question 102 Marks
A bus decreases its speed from 80km/h−1 to 60km/h−1 in 5s. Find the acceleration of the bus.
Answer
Initial speed of the bus, u = 80km/h

$=80\times\frac{5}{18}=22.22\text{m/s}$

Final speed of the bus, v = 60km/h

$=60\times\frac{5}{18}=16.66\text{m/s}$

Time take to decrease the speed, t = 5s

Acceleration, $\text{a}=\frac{\text{v}-\text{u}}{\text{t}}$

$=\frac{16.66-22.22}{5}=-1.112\text{m/s}^2$

Here, the negative sign of acceleration indicates that the velocity of the car is decreasing.

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Question 112 Marks
When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?
Answer
  1. A body is said to have uniform acceleration if it travels in a straight path in such a way that its velocity changes at a uniform rate, i.e., the velocity of a body increases or decreases by equal amounts in an equal interval of time.
  2. A body is said to have non-uniform acceleration if it travels in a straight path in such a way that its velocity changes at a non-uniform rate, i.e., the velocity of a body increases or decreases in unequal amounts in an equal interval of time.
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Question 122 Marks
During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108m/ s−1.
Answer
Speed = 3 × 108 m/ s−1
Time = 5 min = 5 × 60 = 300 secs.
Distance = Speed × Time
Distance = 3 × 108m/ s−1 × 300 secs. = 9 × 1010m.
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Question 132 Marks
Distinguish between speed and velocity.
Answer
Speed
Velocity
Speed is the distance travelled by an object in a given interval of time.
Velocity is the displacement of an object in a given interval of time.
Speed = distance/time
Velocity = displacement/time
Speed is scalar quantity i.e. it has only magnitude.
Velocity is vector quantity i.e. it has both magnitude as well as direction.
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Question 142 Marks
A farmer moves along the boundary of a square field of side 10m in 40s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Answer
https://3.bp.blogspot.com/-PkcnCGLPLBY/VNb22uX4H9I/AAAAAAAADlM/bIHNDiSi0N0/s1600/motion-class-9th-in-text-question-2-1.jpg
Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14m north east from intial position.
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Question 152 Marks
Answer
  1.  

The shaded area which is equal to $\frac{1}{2}\times4\times6=12\text{m}$ represents the distance travelled by the car in the first 4s.

  1.  

The part of the graph in red colour between time 6s to 10s represents uniform motion of the car.

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Question 162 Marks
A driver of a car travelling at $52 km h ^{-1}$ applies the brakes Shade the area on the graph that represents the distance travelled by the car during the period.
(a) Which part of the graph represents uniform motion of the car?
Answer
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Question 172 Marks
Joseph jogs from one end A to the other end B of a straight 300m road in 2 minutes 30 seconds and then turns around and jogs 100m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Answer

Total Distance covered from AB = 300m

Total time taken = 2 × 60 + 30s

=150s

Therefore, Average Speed from AB = Total Distance/Total Time

$=\frac{300}{150}\text{m s}^{-1}$

$=2\text{m s}^{-1}$

Therefore, Velocity from AB = Displacement AB/Time

$=\frac{300}{150}\text{m s}^{-1}​​​​​​​$

$=2\text{m s}^{-1}​​​​​​​$

Total Distance covered from AC = AB + BC

= 300 + 200m

Total time taken from A to C = Time taken for AB + Time taken for BC

= (2 × 60 + 30) + 60s

= 210s

Therefore, Average Speed from AC = Total Distance/Total Time

$=\frac{400}{210}\text{m s}^{-1}​​​​​​​$

$=1.904\text{m s}^{-1}​​​​​​​$

Displacement (S) from A to C = AB – BC

= 300 - 100m

= 200m

Time (t) taken for displacement from AC = 210s

Therefore, Velocity from AC = Displacement (s)/Time(t)

$=\frac{200}{210}\text{m s}^{-1}​​​​​​​$

$=0.952\text{m s}^{-1}​​​​​​​$

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Question 182 Marks
An artificial satellite is moving in a circular orbit of radius 42250km. Calculate its speed if it takes 24 hours to revolve around the earth.
Answer
Radius of the circular orbit, r= 42250km
Time taken to revolve around the earth, t= 24 h
Speed of a circular moving object,
$\text{v}=\frac{(2\pi\text{r})}{\text{t}}$
$=\frac{[2× (22/7)×42250 × 1000]}{(24 × 60 × 60)}$
$=\frac{(2×22×42250×1000)}{(7 ×24 × 60 × 60)}\text{m s}^{-1}$
$=3073.74\text{m s}^{-1}$
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Question 192 Marks
A bus increases its speed from 36km/h to 5 km/h in 10s. Find its acceleration.
Answer
Here u = 36km/hr = 10m/s

v = 54km/hr

= 15m/s

a = ?

t = 10s

$\text{a}=\frac{\text{v}-\text{u}}{\text{t}}$

$\text{a}=\frac{15-10}{10}$

$\text{a}=\frac{15}{10}$

$\text{a}=0.5\text{m/s}^{2}$

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Question 202 Marks
The tip of seconds’ hand of a dock takes 60 seconds to move once on the circular dial of the clock. If the radius of the dial of the clock be 10.5cm, calculate the speed of the tip of the seconds’ hand of the clock. $\Big(\text{Given}\pi=\frac{22}{7}\Big).$
Answer
The speed of a body moving along a circular path is given by the formula:

$\text{v}=\frac{2\pi\text{r}}{\text{t}}$

Given, t = 60 sec

Radius, r = 10.5cm = 0.105m

$\text{v}=\frac{2\pi\text{r}}{\text{t}}=\frac{2\times22\times0.105}{7\times60}=\frac{4.62}{420}=0.011\text{m/s}$

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Question 212 Marks
An object is moving along a straight line with uniform acceleration. The following table gives the velocity of the object at various instants of time:
Time (in sec.):
0
1
2
3
4
5
6
Velocity (in m/s):
2
4
6
8
10
12
14
Plot the graph.
From the graph,
  1. Find the velocity of the object at the end of 2.5 seconds.
  2. Calculate the acceleration.
  3. Calculate the distance covered in the last 4 seconds.
Answer
$\text{a}=\frac{\text{v}-\text{u}}{\text{t}}$

$\text{a}=\frac{4-2}{1}=2\text{m/s}^2$

Now for,

$\text{t}=2.5$

$\text{a}=\frac{\text{v}-\text{u}}{\text{t}}$

$2=\frac{\text{v}-2}{2.5}$

$5=\text{v}-2$

$\text{v}=7$

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Question 222 Marks
What is meant by:
  1. Average speed
  2. Uniform speed?
Answer
  1. Average Speed: The average speed of a body is the total distance travelled divided by the total time taken to cover this distance.

  2. Uniform Speed: A body has a uniform speed if it travels equal distances in equal intervals of time, no matter how small these intervals may be.

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Question 232 Marks
Study the velocity-time graph and calculate.

  1. The acceleration from A to B.
  2. The acceleration from B to C.
  3. The distance covered in the region ABE.
  4. The average velocity from C to D.
  5. The distance covered in the region BCFE.
Answer

  1. $\text{a}=\frac{(25-0)}{(3-0)}=8.3\text{m/s}^{2}$

  2. $\text{a}=\frac{(20-25)}{(4-3)}=5\text{m/s}^{2}$

  3. Distance = Area of triangle ABE.

  4. $=\frac{1}{2}\times3\times25=37.5\text{m}$

  5. $\text{V}=\frac{(20-0)}{2}=10\text{m/s}$

This Distance = Area of trapezium BCFE.

$=\frac{1}{2}(25+20)\times(4-3)=22.5\text{m}$

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Question 242 Marks
After studying the motion of a ball rolling on a straight line as shown in the figure.

Find its distance and displacement covered.
  1. When it rolls from P to Q.
  2. Finally comes back to P (i.e., P to P)(take, P as reference point).
Answer
  1. Distance would be PQ + QR = 12m whereas displacement is path PR= 8m as displacement is the shortest distance between two points.
  2. In this case displacement is zero as initial and final point is same but distance would be PQ + QP = 20m.
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Question 252 Marks
Plot the v-t graph for the following data and state the nature of motion the body:
v(m/s):
0
5
8
12
18
25
30
35
40
45
t(s):
2
4
6
8
10
12
14
16
18
20
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Question 262 Marks
If on a round trip you travel 6km and then arrive back home:
  1. What distance have you travelled?
  2. What is your final displacement?
Answer
  1. Distance travelled = 6km.
  2. Displacement = zero (since final position is same as initial position).
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Question 272 Marks
A person moves a distance of 3km towards east, then 2km towards north and 3.5km towards east. Find:
  1. Distance covered by the person.
  2. Displacement.
Answer
  1. Given that,

A person moves a distance of 3km towards east then 2km towards north and 3.5km towards east.

The distance is

d = 3km + 2km + 3.5km

d = 8.5km

  1. According to figure,

Let a person journey start from A and end at D.

The resultant displacement AD is

Which is the hypotenuses of right angled triangle AED

Here, CD = BE = 3.5km

BC = ED = 2km

Using Pythagorean theorem

$\text{AD}=\sqrt{(\text{AE})^2+(\text{ED})^2}$

$\text{D}=\sqrt{(6.5)^2+(2)^2}$

$\text{D}=6.8\text{m}$

The displacement is 6.8km

The distance and displacement is 8.5km and 6.8km.

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Question 282 Marks
Draw a velocity-time graph to show the following motion:
A car accelerates uniformly from rest for 5s ; then it travels at a steady’ velocity for 5s.
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Question 302 Marks

A body with an initial velocity x moves with a uniform acceleration y. Plot its velocity-time graph.

Answer
We have to plot a graph between velocity and time. From the graph we can conclude that the curve is a straight line showing uniform acceleration.

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Question 332 Marks

A body is moving along a circular path of radius R. What will be the distance travelled and displacement of the body when it completes half a revolution?

Answer
Distance travelled in half a rotation of a circular path is equal to the circumference of semi-circle, i.e.
Distance travelled in half a rotation of a circular path is equal to the circumference of semi-circle, i.e $=\pi\text{R}.$
Displacement = diameter of circle = 2R.
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Question 352 Marks

What can you say about the motion of a body if:

  1. Its displacement-time graph is a straight line?
  2. Its velocity-time graph is a straight line?
Answer
  1. The body has uniform velocity if its displacement-time graph is a straight line. If the straight line is parallel to the time axis then the magnitude of uniform velocity is zero.
  2. The body has a uniform acceleration if its velocity-time graph is a straight line. If the straight line is parallel to the time axis then the magnitude of uniform acceleration is zero.
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Question 362 Marks
How will you show that the slope of the displacement-time graph gives the velocity of the body?
Answer

The adjoining figure shows the displacement-time graph for a body moving with unifrom velocity. clearly. it covers distance s1 and s2 at times t1 and t2 respectively.

Slop of line $\text{PQ}=\text{tan}\theta=\frac{\text{QR}}{\text{PR}}$

$=\frac{\text{s}_{2}-\text{s}_{1}}{\text{t}_{2}-\text{t}_{1}}=\frac{\text{Displacement}}{\text{Time}}$

$\text{As}\frac{\text{Displacement}}{\text{Time}}$ is velocity, so the slope of the distance-time graph gives velocity of the body.

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Question 382 Marks
Bus X travels a distance of 360km in 5 hours whereas bus Y travels a distance of 476km in 7 hours. Which bus travels faster?
Answer
For bus X,

$\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

Speed $=\frac{360}{5}=72\text{km}/\text{h}$

For bus Y,

$\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

Speed $=\frac{476}{7}=68\text{km}/\text{h}$

Speed of bus X is more than that of bus Y. Hence, bus X travels faster.

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Question 392 Marks
On a 120km track a train travels the first 30km at a uniform speed of 30km/h. Calculate the speed with which the train should move rest of the track so as to average 60km/h for the entire trip.
Answer
Time That Should Be Taken So As.

To Attain An

Average Speed Of 60km/hr $=\frac{120}{60}=2\text{ Hours}$

Time Taken To Cover 30km $=\frac{30}{30}=1\text{ Hour}$

Remaining Time = 2 - 1 = 1 Hour

Now Distance Remaining = 90km

Speed $=\frac{\text{Distance}}{\text{Time}}$

$=\frac{90}{1}=90\text{km/hr}$

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Question 402 Marks
In the figure below is shown the time-distance graph of cyclist.

Find out from the graph average speed in the whole journey.
Answer
Average Speed $=\text{Initial}+\frac{\text{Final speed}}{2}$
Average Speed $=0+\frac{4}{2}$
Average Speed = 20M/S.
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Question 412 Marks
The maximum speed of a train is 80km/h. It takes 10h to cover a distance of 400km. Find the ratio of its maximum speed to its average speed.
Answer
According to question the we can use the formula of average speed 
average speed.

$=\frac{\text{total distance}}{\text{total times}}$

$=\frac{400}{10}=40\text{km/hr}$

the ratio of maximum speed to average speed

$=\frac{80}{40}=2$ (as maximum speed of the train is 80km/hr given)

$=2:1$

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Question 422 Marks
If a bus travelling at 20m/s is subjected to a steady deceleration of 5m/s2, how long will it take to come to rest?
Answer
Deceleration, a = -5m/s2

Initial velocity, u = 20m/s

Final velocity, v = 0m/s

t = ?

$\text{a}=\frac{\text{v}-\text{u}}{\text{t}}$

$-5=\frac{0-20}{\text{t}}$

$\text{t}=\frac{20}{5}=4\text{s}$

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Question 432 Marks
A bus running at a speed of 18km/h is stopped in 2.5 seconds by applying brakes. Calculate the retardation produced.
Answer
Initial velocity, u = 18km/h

$\text{u}=18\times\frac{1000}{3600}=\frac{18000}{3600}\text{m/s}=\text{m/s}$

Final velocity, v = 0m/s

Time, t = 2.5 sec

Acceleration, a = ?

Using, v = u + at

$\text{a}=\frac{\text{v-u}}{\text{t}}=\frac{0-5}{2.5}2\text{m/s}^2$

So, retardation is 2m/s2.

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Question 442 Marks
A ball hits a wall horizontally at 6.0ms-1. It rebounds horizontally at 4.4ms-1. The ball is in contact with the wall for 0.040s. What is the acceleration of the ball?
Answer
Initial velocity, u = 6m/s
Final velocity, v = -4.4m/s (the ball rebounds in opposite direction)
Time, t = 0.040s
Acceleration velocity $=\frac{\text{u}-\text{v}}{\text{t}}=\frac{-4,4-6}{0.040}=-260\text{m/s}^2$
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Question 452 Marks
A train starting from stationary position and moving with uniform acceleration attains a speed of 36km per hour in 10 minutes. Find its acceleration.
Answer
Initial velocity, u = 0m/s

Final velocity, v = 36km/h = 10m/s

Time, t = 10min = 10 x 60 = 600 sec

 Acceleration, a = ?

 Acceleration $=\frac{\text{Final velocity-Initial velocity}}{\text{time taken}}$

So, $\text{a}=\frac{\text{v-u}}{\text{t}}$

$=\frac{10-0}{600}=\frac{10}{600}\text{m/s}^2$

$=\frac{1}{60}\text{m/s}^2-0.16\text{m/s}^2$

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Question 462 Marks
A train starting from Railway Station attains a speed of 21m/s in one minute. Find its acceleration.
Answer
u = 0m/s

v = 21m/s

Time, t = 1min = 60sec

$\text{a}=\frac{\text{v}-\text{u}}{\text{t}}$

$\text{a}=\frac{\text{21}-\text{0}}{\text{60}}$

$\text{a}=\frac{21}{60}=0.35\text{m}/\text{s}^2$

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Question 472 Marks
An aircraft travelling at 60km/h accelerates steadily at 10km/h per second. Taking the speed of sound as 1100km/h at the aircraft’s altitude, how long will it take to reach the ‘sound barrier’?
Answer
Initial velocity, u = 600km/h

Final velocity, v = 1100km/h

Acceleration = 10km/h/s = 600km/h2

From relation, $\text{a} = \frac{(\text{v-u})}{\text{t}}$

$\text{t} = \frac{(\text{v-u})}{\text{a}}$

$\text{t} = \frac{(1100-600)}{600} = \frac{500}{600}= \frac{5}{6}\text{hr} = 50 \text{sec}$

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Question 482 Marks
A motorcycle moving with a speed of 5m/s is subjected to an acceleration of 0.2m/s2. Calculate the speed of the motorcycle after 10 seconds, and the distance travelled in this time.
Answer
Initial velocity, u = 5m/s

Final velocity, v = ?

Acceleration, a = 0.2m/s2

Time, t = 10 sec

Using, v = u + at

v = 5 + 0.2 x 10

v = 5 + 2 = 7m/s

Now distance travelled in time is calculated;

Using, $\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$

$\text{s}=5\times10+\frac{1}{2}\times0.2\times10\times10$

$\text{s}=50+10=60\text{m}$

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Question 492 Marks
Arrange the following speeds in increasing order (keeping the least speed first):
  1. An athlete running with a speed of 10m/s.
  2. A bicycle moving with a speed of 200m/min.
  3. A scooter moving with a speed of 30km/h.
Answer
Speed of athelete = 10m/s
Speed of bicycle = 200m/min $= \frac{200}{60}\text{m/s} = 3.33\text{m/s}$
Speed of scooter = 30km/h $= \frac{30000}{3600}\text{m/s} = 8.33\text{m/s}$
3.33m/s < 8.33m/s < 10m/s
i.e. 200m/min < 30km/h < 10m/s
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Question 502 Marks
A motorcyclist starts from rest and reaches a speed of 6m/s after travelling with uniform acceleration for 3s. What is his acceleration?
Answer
intial velocity = 6m/s

Time = 3sec

Acceleration $=\frac{\text{Final velocity - Intial velocity}}{\text{Time taken}}$

$=\frac{(6-0)}{3}=\frac{6}{3}=2\text{m/sec}^2$

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