Motion in a Straight Line Test 13

As the body starts from rest,
∴ u = 0
Let a be constant acceleration of the body.
Distance travelled by the body in (t/2) s is

Distance travelled by the body in t s is

Free fall of an object in vacuum is a case of motion with uniform acceleration.

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.

Distance travelled by a body in nth second is

Here, u = 0, a = g
∴ Distance travelled by the body in 1st second is

Distance travelled by the body in 2^nd second is

Distance travelled by the body in 3^rd second is

and so on.
Hence, the distance covered by a freely falling body in its first, second, third ..... nth second of its motion forms an arithmetic progression.

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

At the highest point velocity of the ball becomes zero, but its acceleration is equal to g.

Let a be constant acceleration of auto.
Here, u = 30ms⁻¹, v = 50ms⁻¹, S = 180 m
As v² − u² = 2aS

Solving this quadratic equation by quadratic formula, we get = 4.5s, −18 s, (t can't be negative)
∴ t = 4.5 s

Suppose the body passes the upper point at t second and lower point at (t + 1) s, then

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.

Here, u = 0, g = 10 ms^-2, h = 1 km = 1000 m
As v² - u² = 2gh
∴ v² = 2gh

For first stone,
taking the vertical upwards motion of the first stone up to highest point
Here, u = u₁, v = 0 (At highest point velocity is zero)
a = -g, S = h₁
As v² − u² = 2aS

For second stone,
Taking the vertical upwards motion of the second stone up to highest point
here, u = U², v = 0, a = −g, S = h²
As v² − u² = 2as

As per question

Subtract (ii) from (i), we get,

On substituting the given information, we get

or u₁ = 20ms⁻¹ and u₂ = U₁/2 = 10ms^-1

Taking vertical upward motion of the ball upto highers point
Here, u = 20 ms^-1
v = 0 (at highest point velocity is zero)
a = -g = -10ms^-2
As v² = u² + 2as
0 = (20)² + 2(-10)(S) or 

Let t₁ be the time taken by the ball to reach the highest point.
here, v = 0, u = 20ms⁻¹, a = −g = −10ms⁻², t = t₁
As v = u + at
∴ 0 = 20 + (−10)t₁ or t₁ = 2s
Taking vertical downward motion of the ball from the highest point to ground.
Here, u = 0, a = +g = 10ms⁻², S = 20 m + 25 m = 45 m, t = t₂

Total time taken by the ball to reach the ground = t₁ + t₂ = 2s + 3s = 5s

For train B,


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

Here, u = 10ms⁻¹, t = 3s, v = 16ms⁻¹

Now velocity at 2 s, before the given instant
10 = u + 2 x 2
∴ u = 6 m s⁻¹ (since v = u + at)

On solving, u = −6ms⁻¹, a = 4ms⁻²
Distance travelled in next 3 seconds = S₈ − S₅

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

Here, a = g − bv
When an object falls with constant speed v_c, its acceleration becomes zero.
∴ g − bv_c = 0 or v_c = g/b

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)

Let the two stones meet at time t.
for the first stone,

for the second stone,

Displacement is same
∴ S₁ = S₂

When car overtakes motorcycle, both have travelled the same distance in the same time. Let the total distance travelled be S and the total time taken to overtake be t.

For motorcycle:
Maximum speed attained = 36kmh⁻¹

Since its acceleration = 1.0ms⁻², the time t₁ taken by it to attain the maximum speed is given by

The distance covered by motorcycle in attaining the maximum speed is

The time during which the motorcycle moves with maximum speed is (t − 10)s.
The distance covered by the motorcycle during this time is 

∴ Total distance travelled by motorcycle in time t is

For car:
Maximum speed attained =

Since its acceleration = 0.5ms⁻²

The time taken by it to attain the maximum speed is given by
15 = 0 + 0.5 x t₂ or t₂ = 30s (∵ u = 0)
The distance covered by the car in attaining the maximum speed is 

The time during which the car moves with maximum speed is (t − 30)s.
The distance covered by the car during this time is

∴ Total distance travelled by car in time t is

From equations (i) and (ii), we get
10t − 50 = 151 − 225 or 51 = 175 or 1 = 35s

Displacement in last 2 second

Acceleration a = v/n
(∵ t = ns)
Displacement in last 2s =

Time taken by body A, t₁ = 5s

Acceleration of body A = a₁

Time taken by body B, t₂ = 5 − 2 = 3s

Acceleration of body B = a₂

Distance covered by first body in 5th second after its start,

Distance covered by the second body in the 3rd second after its start,

Since S₅ = S₃

For first part of penetration, by equation of motion

For later part of penetration