50 questions · timed · auto-graded
Explanation:
The difference in velocities is increasing with time as both of them have more constant but different acceleration.
Explanation:
Gravity is the only force acting on the stone when it is released. And, we know that gravity is always in the downward direction.
Explanation:
The vertical distance between A and B is of 2 units and the horizontal distance is 1 unit. Hence, to calculate the displacement, we can make a right-angled triangle as shown and can find the distance (AB) by Pythagoras theorem. Distance (AB) = Square root (22 + 12) = Square root (5).
Explanation:
Let the upward direction to be negative.
Initial velocity of stone u = -4m/s
Acceleration of the stone a = g = 9.8m/s2
Velocity of stone v = u + at
$\therefore$ v = -4 + 9.8 × 3 = 25.4m/ s
Explanation:
A object is said to have an acceleration if it changes its velocity either by increasing its speed, decreasing its speed or changing the direction of its velocity.
Since the car and the truck move with constant speed, thus they have zero acceleration.
But the van is slowing down its speed, thus it has deacceleration.
Explanation:
The ball thrown upward will have zero velocity in1s. It returns back to thrown point in another 1s with the same velocity as second.
Thus the difference will be 2s.
Explanation:
The average velocity is obtained by dividing total displacement by total time taken.
Instantaneous velocity is calculated at an instant and not over a period of time.
Speed is distance divided by time. Velocity is said to be uniform when velocity at every instant is equal to the average velocity.
Explanation:
Since all the balls have have zero velocity initially and they experience equal acceleration due to gravity and travel equal distance. Thus all the balls take equal time to reach the ground.
Explanation:
At the moment when the car will be overtaking the truck their velocitis will be same.
So suppose after t second from the start of the motion the car overtake the truck then the velocity of the car at that moment will be v = 0 + at = t × 1m/ s2 = tm/ s and the velocity of the truck at that moment will be u = 6m/ s(constant) both should be same i.e u = v or 6 = t or t = 6 second so the distance covered by truck oe car because both will be at same place ast that point of time in this time will be s = vt = 6 × 6 = 36 meter
Explanation:
The correct answer is d2 = (x2 - x1)2 + (y2 - y1)2. This expression can be found out using Pythagoras theorem in Cartesian coordinate system.
Build a right-angle triangle with sides parallel to the axes and the hypotenuse joining the two points to construct the right-angle triangle.
Explanation:
Displacement in 10 seconds is $\text{ut}+\frac{\text{at}^2}{2}$
$=5\times10+2\times\frac{10^2}{2}=150\ \text{meter}$
Final position is initial position + displacement 5 + 150 = 155meter
Explanation:
Let u and v be the speed of train and wind respectively. The speed of steam track of train moving in the direction of wind = u - v.
The speed of steam track of train moving in the opposite direction of wind = u + v
As per question, (u + v) = 2(u - v)
u = 3v
Explanation:
The accelerated motion of a body changes due to a change in speed, direction of motion, velocity.
As acceleration posses magnitude and direction. Its magnitude changes by a change in speed, velocity, and direction can be changed by the direction of motion and velocity.
Explanation:
The displacement covered in the first five seconds can be obtained by putting t = 5 in the equation. Therefore, x = 55 + 2(5) = 25 + 10 = 35. Hence the answer is 35 units.
Explanation:
Key concept: The time rate of change of velocity of an object is called acceleration of the object.
It is a vector quantity. Its direction is same as that of change in velocity (Not of the velocity).
In the table: Possible ways of velocity change.
| When only direction of velocity changes. | When only magnitude of velocity changes. | When both magnitude and direction of velocity change. |
| Acceleration perpendicular to velocity. | Acceleration parallel or antiparallel to velocity. | Acceleration has two components-one is perpendicular to velocity and another parallel or antiparallel to velocity. |
| E.g.: Uniform circular motion | E.g.: Motion under gravity. | E.g.: Projectile motion. |
Here we will take upward direction positive. As. the lift is coming in downward direction, the displacement will be negative. We have to see whether the motion is accelerating or retarding.
We know that due to downward motion displacement will be negative. When the lift reaches 4th floor and is about to stop velocity is decreasing with time, hence motion is retarding in nature. Thus, x < 0; a > 0.
Asdisplacementisinnegativedirection, velocity will also be negative, i.e. v < 0.
The motion of lift will be shown like this.
Explanation:
Key concept: Instantaneous velocity : Instantaneous velocity is defined as the rate of change of position vector of particles with time at a certain instant of time.
Instantaneous velocity $\vec{\text{v}}=\DeclareMathOperator*{\median}{\text{lim}} \median\limits_{\Delta\text{t}\rightarrow0}\frac{\Delta\vec{\text{r}}}{\Delta\text{t}}=\frac{\text{d}\vec{\text{r}}}{\text{dt}}$
Instantaneous acceleration $=\vec{\text{a}}=\DeclareMathOperator*{\median}{\text{lim}} \median\limits_{\Delta\text{t}\rightarrow0}\frac{\Delta\vec{\text{v}}}{\Delta\text{t}}=\frac{\text{d}\vec{\text{v}}}{\text{dt}}$
By definition $\vec{\text{a}}=\frac{\text{d}\vec{\text{v}}}{\text{dt}}=\frac{\text{d}^2\vec{\text{x}}}{\text{dt}^2}\Big[\text{As}\ \vec{\text{v}}=\frac{\text{d}\vec{\text{x}}}{\text{dt}}\Big]$
i.e., if x is given as a function of time, second time derivative of displacement gives acceleration.
In such type of problems we have to analyze whether the motion is accelerating or retarding. When acceleration is parallel to velocity, velocity of particle increases with time, i.e. motion is accelerated. And when acceleration is anti-parallel to velocity, velocity of particle decreases with time, i.e. motion is retarded. During retarding journey, particle will stop in between.
According to the problem, displacement of the particle is given as a function of time.
$\text{x}=(\text{t}-2)^2$
By differentiating this equation w.r.t. time we get velocity of the particle as a function of time.
$\text{v}=\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{t}-2)^2=2(\text{t}-2)\text{m/s}$
If we again differentiate this equation w.r.t. time we will get acceleration of the particle as a function of time.
Acceleration, $\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}}{\text{dt}}[2(\text{t}-2)]$
$=2[1-0]=2\text{m/s}^2$
When $\text{t}=0;\ \ \text{v}=-4\text{m/s}$
$\text{t}=2\text{s};\ \ \text{v}=0\text{m/s}$
$\text{t}=4\text{s};\ \ \ \text{v}=4\text{m/s}$
That means particle starts moving towards negative axis, then at = 0, with a speed 4m/s, at t = 2 it stops and start coming backward. At t = 4 its speed is +4m/s.
v - t graph is shown in graph (a) and speed-time graph of the same situation is shown in graph (b).
Distance travelled = Area of the speed-time graph
= area OAC + area ABD
$=\frac{4\times2}{2}+\frac{1}{2}\times2\times4=8\text{m}$
Explanation:
Let s be the height of a particular point where the ball crosses in time t1 and t2 seconds while going upwards and coming downwards. If u is the initial velocity of projection of ball, then
$\text{s}=\text{ut}_1-\frac{1}{2}\text{gt}^2_1=\text{ut}_2-\frac{1}{2}\text{gt}^2_2$
$\text{u}(\text{t}_2-\text{t}_1)=\frac{1}{2}\text{g}(\text{t}^2_2-\text{t}_1)$
$\text{u}=\frac{1}{2}\text{g}(\text{t}_2+\text{t}_1)$
If T is the time taken by ball to reach to its highest point then using the relation v = u + at, we have 0 = u + (-g)T
$\text{T}=\frac{\text{u}}{\text{g}}=\frac{1}{2}\frac{\text{g}(\text{t}_2+\text{t}_1)}{\text{g}}$
$=\frac{1}{2}(\text{t}_2+\text{t}_1)$
Explanation:
As the bullet is fired vertically upwards, thus the bullet does not have velocity in horizontal forward direction and hence it has zero horizontal displacement.
Also the cart has acceleration in forward direction, thus the cart has finite displacement in forward direction.
Hence the bullet goes up and lands behind the cart.
Explanation:
Velocity is uniform in both cases; that is, acceleration is zero.
$\text{x = v}_1\text{t}_1\Rightarrow\text{t}_1=\frac{\text{x}}{\text{v}_1}$
$\text{x = v}_2\text{t}_2\Rightarrow\text{t}_2=\frac{\text{x}}{\text{v}_2}$
Total displacement, $\text{x}'=\text{2x}$
Total time, $\text{t}=\text{t}_1+\text{t}_2$
$\therefore$ Average velocity, $\text{v}=\frac{\text{x}'}{\text{t}}=\frac{2\text{v}_1\text{v}_2}{\text{v}_1+\text{v}_2}$
$\Rightarrow\frac{2}{\text{v}}=\frac{1}{\text{v}_1}+\frac{1}{\text{v}_2}$
Explanation:
When the body is moving with terminal velocity, the velocity does not change. It means that equal displacement is being covered in equal time intervals. Hence the average velocity remains constant.
Explanation:
Since, average velocity, $\text{v}=\frac{\Delta \text{x}}{\Delta \text{t}}=\frac{\text{Displacement}}{\text{Time interval}}$
Thus, average velocity depends on the displacement and hence it depends on the sign of the displacement.
Explanation:
We know that distance traveled by an object is the area under it speed time graph.
Now, in this case, as the area under the speed-time graph is increasing from 0 - 10 minutes.
So, the distance will keep on increasing from 0 - 10 minutes.
Explanation:
Here, initial velocity is u = 0m/ s (as airplane starts from rest)
Final velocity is v = 80m/ s and distance traveled, S = 1000m.
Let a be the required acceleration.
Using formula v2 − u2 = 2aS,
(80)2 − 02 = 2a(1000)
So, we get a = 3.2m/ s2
Explanation:
When the body is in the state of rest, there is no motion. Hence there is no kinetic energy, hence the total energy of the body is stored as its potential energy. The total energy is the sum of kinetic and potential energies.
Explanation:
Take any point on the circular path. The car moves in a circle and hence will come back to the same point after a definite time interval.
Therefore, the displacement is 0. Hence, the average velocity is zero.
Explanation:
The kinematic equations for uniform acceleration do not apply in case of uniform circular motion because in this case the magnitude of acceleration is constant but its direction is changing.
Explanation:
The toy train will speed up if the rate of change of velocity will increases with respect to time.
Therefore, the train will speed up in the intervals 0 to A and D to E.
Explanation:
When the ball rises up, there will be a point where it will be in the state of instantaneous rest. At the this position the speed of the ball will be 0. Speed is maximum at the initial and final points.
Explanation:
As velocity and acceleration are in opposite directions, velocity will become zero after some time (t) and the particle will return.
$\therefore0=\text{u}-\text{at}$
$\Rightarrow\text{t}=\frac{\text{u}}{\text{a}}$
Because the value of acceleration is not given, we cannot say that the particle will return after/ before 10 seconds.
Explanation:
Acceleration is defined as the rate of change of velocity.
acceleration a $=\frac{\text{change in velocity}}{\text{time interval}}$
Hence, any change in the speed or direction would cause a change in velocity, and hence will produce acceleration.
Explanation:
Average speed is the total distance divided by total time taken.
Total displacement (d = vt) = 20 × 20 + 15 × 10 + 10 × 5 = 600m.
Total time = 20 + 10 + 5 = 35s.
Therefore, average speed $=\frac{600}{35}=17.14\text{m}/\ \text{s}.$
Explanation:
The gradient or slope of any graph tells us the value of its differential at that point.
Since acceleration is the differential of velocity with respect to time, the gradient is equal to the acceleration.
Explanation:
Taking upward motion of balloon for 8 seconds; we have
u = 0; a = 1.25m/s2; t = 8 s; v = ?; s = ?.
Here v = u + at = 0 + 1.25 × 8 = 10m/s
$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
$=0+\frac{1}{2}\times1.25\times8^2=40\text{m}$
Taking downward motion of released stone from balloon at height 40m we have,
a = -10m/s; a = 10m/s2; s = 40m; t = ?
As, $\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
so, $40=-10\text{t}+\frac{1}{2}\times10\times\text{t}^2$
or t2 - 2t - 8 = 0 On solving t = 4s.
Explanation:
Let t second be the time of flight of the first body after meeting, then (t − 4) second will be the time of flight of the second body.
Since, h1 = h2
$\therefore98\text{t}-\frac{1}{2}\text{gt}^2=98(\text{t}-4)\text{g}(\text{t}-4)^2$
On solving, t = 12s.