Test: Kinematic Equations for Uniformly Accelerated Motion (April 18) - JEE MCQ

S = vt + 1/2at²
It is not a kinematic equation of motion.
All others are three kinematic equations of motion.

The sign of acceleration does not tell us whether the particle's speed is increasing or decreasing. The sign of acceleration depends on the choice of the positive direction of the axis.
For example: If the vertically upward direction is chosen to be positive direction of the axis, the acceleration due to gravity is negative. If a particle is falling under gravity, this acceleration though negative results in increase in speed.

As distance travelled in successive seconds differ by 2 m each, therefore acceleration is constant = 2ms⁻²

there will be acceleration dur to friction

Let t be time taken by the ball to reach the highest point.
As v = u + at
Here, u = 30ms⁻¹
v = 0 (At highest point velocity is zero)
a = - g = −10ms⁻²
∴0 = 30 - 10t or t = 3 s
∴ Time taken by the ball to return to player's hand
= 3 s + 3 s = 6 s.

When the lift starts moving upwards with uniform speed of 5 ms⁻¹, there is no change of relative velocity of ball w.r.t. girl. Hence, even in this case the ball will return to the girl’s hands in the same time i.e 10s.

For train B,


Original distance between A and B = SB − SA = 2250m − 1000m = 1250m

Let d_s is the distance travelled by the vehicle before it stops.

Here, final velocity v = 0, initial velocity = u, S = d_s

Using equation of motion

Let t₁, t₂, t₃ be the timings for three successive equal heights h covered during the free fall of the particle. Then

Subtracting (i) from (ii), we get

From (i),(iv) and (v), we get
t₁:t₂:t₃ = 1:(√2-1) : (√3-√2)