Question

A car is travelling with a speed of 36km/hr. The driver applies the break and returns the car uniformily . The car is stopped in 5 sec. Find the return time of tha car and distance travelled before it is stopped after applying the break

Answer 1

Given :-

u = 36 km/hr

= 36 × 5/18

= 10 m/s

t = 5 s

v = 0

To find :-

The retardation time of the car.

And distance travelled after applying the break.

Solution:-

After applying the breaks the final velocity of the car becomes 0.

By using first eq. of motion :-

→$V = u + at$

→$0 = 10 + a \times 5$

→$0 = 10 +5a$

→$5a = -10$

→$a = \dfrac{-10}{5}$

→$a = -2 m/s^2$

Now,

Distance travelled by car after applying the breaks.

→$2as = v^2 - u^2$

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

→$-4s = -100$

→$s = \dfrac{-100}{-4}$

→$s = 25 m$

Hence,

Distance travelled by the car Will be 25 m.

Note :- Please check your question again .

Return time = Retardation

Answer 2

Answer:

Distance travelled by car after applying brakes = 25 meter

Step by step explanations :

given that,

A car is travelling with a speed of

36km/hr ⛟

and

The driver applies the break and returns the car uniformily .and it stopped in 5 seconds

here,

we have,

initial velocity of the car = 36 km/h

= 36 × 1000m/3600

= 10 m/s

final velocity of the car = 0 m/s.

[car stops by applying brakes]

time taken to stop = 5 seconds

let the distance travelled by the car after applying brakes be s

now we have,

initial velocity of car(u) = 10 m/s

final velocity(v) = 0 m/s

time taken(t) = 5 seconds

distance travelled(s) = s

so,

by the equation of motion,

➩ v = u + at

putting the values,

0 = 10 + a × 5

5a = -10

a = -10/5

a = -2 m/s²

now,

again by the equation of motion,

➩ v² = u² + 2as

putting the values,

(0)² = (10)² + 2(-2)s

-4s = -100

S = -100/-4

S = 25 m

so,

Distance travelled by car after applying brakes

= 25 meter

___________________

Extra information :

➩ Negation of Acceleration shows the retardation