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Q.40. A car is travelling with a speed of 36 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?

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Solution

Dear student,

$I n i t i a l v e l o c i t y u = 36 k m / h = 10 m / s f i n a l v e l o c i t y = 0 t i m e t a k e n = 5 s l e t a i s t h e a c c e l e r a t i o n v = u + a t 0 = 10 + 5 a a = - 2 m / s^{2} r e t a r d a t i o n i s - 2 m / s^{2} d i s \tan c e t r a v e l l e d i s : s = u t + \frac{1}{2} a t^{2} s = 10 \times 5 - \frac{1}{2} \times 2 \times 5 \times 5 s = 25 m$
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Solve this: Q.40. A car is travelling with a speed of 36 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, Initial velocity u = 36 km/h = 10 m/sfinal velocity = 0time taken = 5 slet a is the accelerationv = u + at0 = 10 + 5aa =-2 m/s2retardation is -2 m/s2distance travelled is:s = ut + 12at2s= 10×5-12×2×5×5s = 25 m Regards

Solve this:
Q.40. A car is travelling with a speed of 36 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?