Question

only √√√√physics experts can solve...........,.......... questions... A car moves linearly with uniform retardation. If the
car covers 40 m in the last 2 seconds of its motion, what is the velocity of car at the beginning of the last second​

Answer 1

Answer:

The acceleration of the car is given to be uniform so we can straight away use the kinematical equations here.

Let us say that it's initial Velocity was

u.

So,

this means from the first and third kinematical equation.

u = a(2)

and,

u^2 = 2a(40)

So,

u^2 - 40u = 0

u = 40 m/s

neglecting the solution that u is 0 m/s which it can't be otherwise motion wouldn't occur.

So, now

Acceleration = -(u/2) = -20 m/s^2

So,

Velocity at the last second = 40-20 m/s = 20 m/s

So, the answer is 20 m/s

Thank you Shubendu8898 sir ! I apologise for my bragging statements in the previous answer !

Answer 2

Solution

• Initial Velocity Of The Car is 80 m/s

Given

• Distance Covered (s) = 40 m
• Time Taken (t) = 2 seconds

To finD

• Velocity of the car at beginning of last second

The car is undergoing retardation,thus v = 0 m/s

From the third equation of motion,

$\sf \: {v}^{2} - {u}^{2} = 2as \\ \\ \dashrightarrow \: \sf \: 0 - u^2 = 2(- a)s \\ \\ \dashrightarrow \sf a = \dfrac{u^2}{2s}---------------(1)$

Velocity at last second :

• Time (t) = 1 s
• Retardation is uniform
• Final Velocity (v) = 0 m/s

$\sf v = u + at \\ \\ \longrightarrow \sf 0 = u + \dfrac{u^2}{2s}t \\ \\ \longrightarrow \sf -2us = u^2t \\ \\ \longrightarrow \sf u = - \dfrac{2s}{t} \\ \\ \longrightarrow \boxed{\boxed{\sf u = - 80 m/s }}$