Question

When brakes are applied to a car moving with a uniform velocity of 36 km/hr, it comes to
stop after covering a distance of 50 m. Find the retardation of the car
(A) 1 m/s2
(B) 2 m/s2
(C)
= m/s
2
(D)
3 m/s2
Os
Find the time taken by the car to come to a stop in the above question.
(A) 5s
(B) 8s (C) 10s
(D)20 s​

Answer 1

Answer:-

1) a = -1m/s²

Option →A

2) t = 10 s

Option → C

Given :-

u = 36 km/hr

= 36 × 5/18

= 10 m/s

s = 50 m

v = 0

To find :-

The retardation of the car .

And the time taken to stop.

Solution:-

Retardation of the car will be given by :-

$\huge \boxed{2as = v^2 - u^2}$

→$2 \times a \times 50 = (0)^2 - (10)^2$

→$100a = -100$

→$a =\dfrac{-100}{100}$

→$a = -1 m/s^2$

hence,

Retardation of the car is -1 m/s².

The time taken to stop the car is given by :-

$\huge \boxed {v = u + at}$

→$0 = 10 -1 \times t$

→$-10 = -t$

→$t = 10 s$

hence,

Time takes to stop the car is 10 s.

Answer 2

Velocity = 36km/hr

Or, 36*1000/3600 = 10m/s

It comes to a stop after covering 50m

Retardation is a negative acceleration.

Thus, final velocity= 0

initial velocity = 10m/s

2as = v^2-u^2

a = 0-100/100

a = -1m/s

Time:

V = u+at

0 = 10+(-1*t)

0 = 10-t

10-t=0

10=t