A car accelerates from rest at a constant rate for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t, the maximum velocity acquired by the car is given by

\(\left ( \frac{\alpha ^{2}+\beta ^{2}}{\alpha \beta } \right )t\)
\(\left ( \frac{\alpha ^{2}+\beta ^{2}}{\alpha \beta } \right )t\)
\(\left ( \frac{\alpha +\beta }{\alpha \beta } \right )t\)
\(\left ( \frac{\alpha \beta}{\alpha +\beta} \right )t\)

Solution

A parachutist after bailing out falls 50m without friction.When parachute opens, it decelerates at 2 m/s². He reaches the ground with a speed of 3 m/s. At what height, did he bailout ?

182 m
91 m
111 m
293 m

Solution

A stone is just released from the window of a train moving along a horizontal straight track. The stone will hit the ground following a

straight line path
circular path
parabolic path
hyperbolic path

Solution

The horizontal velocity of the stone will be the same as that of the train. In this way, the horizontal motion will be uniform motion. The vertical motion will be controlled by the force of gravity, i. e., vertical motion is accelerated motion. Thus the resultant motion will be along a parabolic trajectory.

The relation between time t and distance x is t = αx²x + βx where α and β are constants. The retardation is

2αv³
2βv³
2αβv³
2β²v³

Solution

A point initially at rest moves along x-axis. Its acceleration varies with time as 2a(6t+5)m/s². If it starts from origin,the distance covered in 2 s is

20 m
18 m
16 m
25 m

Solution

A balloon starts rising from the ground with an acceleration of 1.25 ms^-2. After 8 s, a stone is released from the balloon.The stone will (Taking g = 10 m s^-2)

begin to move down after being released
reach the ground in 4s
cover a distance of 40 m in reaching the ground
will have a displacement of 50 m.

Solution

v =1.25 × 8 ms^-1= 10 ms^-1

s=¹⁄₂ × 1.25 × 8 × 8m= 40m

Now , 40= -10t + ¹⁄₂ × 10 × t²

or 5t² - 10t - 40 = 0

or t² -2t - 8 = 0 or t =4 s.

Two particles A and B are connected by a rigid rod AB. The rod slides along perpendicular rails as shown here. The velocity of A to the left is 10 m/s. What is the velocity of B when angle α= 60°?

9.8 m/s
10 m/s
5.8 m/s
17.3 m/s

Solution

Here, the particle B moves upwards. Let the upward velocity of B be v then \(\frac{v}{10}\) = tan 60°

The distance time graph of a particle at time t makes angles 45° with the time axis. After one second, it makes angle 60° with the time axis. What is the acceleration of the particle?

√3-1
√3+1
√3
1

Solution

Velocity at time t is tan 45° = 1. Velocity at time(t = 1) is tan 60°=√3. Acceleration is change in velocity in=√3-1.

Which of the following graph cannot possibly represent one dimensional motion of a particle?

All of the above

Solution

In (a), at the same time particle has two positions which is not possible.In (b), particle has two velocities at the same time. In (c), speed is negative which is not possible

For the velocity time graph shown in the figure below the distance covered by the body in the last two seconds of its motion is what fraction of the total distance travelled by it in all the seven seconds?

¹⁄₂
¹⁄₄
²⁄₃
¹⁄₃

Solution

Distance in last two second

= ¹⁄₂ × 10 × 2 = 10 m.

Total distance = ¹⁄₂ × 10 × (6 + 2)= 40 m.

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