Question

A car
moves with uniform
velocity of 40 ms for 5 s. It
comes to rest in the next 10 s s
with uniform deceleration. Find
(i) deceleration (ii) total
distance travelled by the car.
(using equations of motion)​

Answer 1

Answer:

(i) deceleration :-

a= v-u/t²

= 40-0/10

= 40/10

= 4

(ii) distance traveled:-

v= s/t¹

s = v×t¹

s=40×5

s=200m

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Answer 2

Answer:

i) 8 m/s²

ii) 100 m

Explanation:

Given

u = 40 m/s

v = 0 m/s

t = 5 s

i) a = (v-u)/t

= (0-40)/5

= -40/5

-8 m/s²

Therefore, the acceleration of the car is -8 m/s². So the deceleration is 8 m/s².

ii) 2as = v² - u²

= s = (v²-u²)/2a

= {(0)²-(40)²}/2(-8)

= -1600/-16

= 100 m

Therefore, the distance covered by the car is 100 m.

Hope it helps!!