Question

a particle starting from rest accelerated for 5sec then the acceleration is ceased if it describes 100m in the next 10sec find the acceleration in the first place and the total distance covered​

Answer 1

Answer:

2m/s^2

Explanation:

In 10 seconds, it travels 100m. This means it is moving at 10m/s when the acceleration has ceased. (100/10 = 10 , as speed = distance/time)

If the particle accelerated at a constant rate, it would have went from 0m/s to 10m/s in 5 seconds. This means it was accelerating at 2m/s^2, as 10/5 = 2.

Answer 2

Answer:

If in next 10seconds particle covered 100metres without any acceleration then it's velocity must be

10m/sec. Now in first 5seconds it's velocity becomes 0 to 10m/sec , then it's acceleration during that time must

be 2metres per second square as a = [v-u]÷t,

(10m/sec - 0 m/sec) ÷5sec = 10 ÷ 5 = 2m/sec^2.

Now 2as = v^2-u^2, so distance covered in first five seconds is this,

2×2m/sec× s = 10^2 - 0^2

4s = 100 - 0

s = 100÷4

s = 25metres

now total distance covered is equal to 100+25=125metres.