A car is travelling with a speed of 72 km/h. The driver applied the brakes and
retards the car uniformly. The car is stopped in 5 sec.
Find (i) The acceleration of car and (ii) Distance before it stops after applying breaks.

Open in App

Solution

Dear student
$72 k m / h r = 72 \times \frac{5}{18} = 20 m / s v = u + a t 0 = 20 - a \times 5 (p a r t i c l e i s r e t a r d e d) a = \frac{20}{5} = 4 m / s^{2} v^{2} - u^{2} = 2 a x 0 - 20^{2} = 2 \times (- 4) \times x - 400 = - 8 \times x x = \frac{400}{8} = 50 m R e g a r d s$

Suggest Corrections

Similar questions

Q. A car is travelling with a speed of 36km/h. The driver applies brakes retards the car uniformly. The car is stopped in 5seconds. Find the retardation of the car and the distance travelled before it is stopped.

Q. A driver of a car travelling at

52

h^{- 1}

applies the brakes and accelerates uniformly in the opposite direction. The car stops in

5

s. Another driver going at

3

h^{- 1}

in another car applies his brakes slowly and stops in

10

s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

A car travels with a uniform velocity of 25 $m s^{- 1}$ for 5 s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 s.Find:(i) the distance which the car travels before the brakes are applied,(ii) the retardation,and (iii) the distance travelled by the car after applying the brakes.

Q. A car is moving in a straight line with speed 18 km/hr. It is stopped in 5s by applying the brakes. Find (i) the retardation and (ii) speed of the car after 2s of applying the brakes.

Join BYJU'S Learning Program

A car is travelling with a speed of 72 km/h. The driver applied the brakes and retards the car uniformly. The car is stopped in 5 sec. Find (i) The acceleration of car and (ii) Distance before it stops after applying breaks.

Dear student 72km/hr=72×518=20m/sv=u+at0=20-a×5 (particle is retarded)a=205=4m/s2v2-u2=2ax0-202=2×(-4)×x-400=-8×xx=4008=50mRegards

A car is travelling with a speed of 72 km/h. The driver applied the brakes and
retards the car uniformly. The car is stopped in 5 sec.
Find (i) The acceleration of car and (ii) Distance before it stops after applying breaks.

Dear student
$72 k m / h r = 72 \times \frac{5}{18} = 20 m / s v = u + a t 0 = 20 - a \times 5 (p a r t i c l e i s r e t a r d e d) a = \frac{20}{5} = 4 m / s^{2} v^{2} - u^{2} = 2 a x 0 - 20^{2} = 2 \times (- 4) \times x - 400 = - 8 \times x x = \frac{400}{8} = 50 m R e g a r d s$