Question

1. A body moving with a speed of 36 km/h
is brought to rest in 10 s. What is the
negative acceleration and the distance
travelled by the body before coming to rest?​

Answer 1

Initial velocity u of the body = 36 km/h =( 36 ×1000/3600)= 10 m/s

Final velocity v of the body = 0 m/s

Time in which the body is brought to rest = 10 s

Using the relation: v -u = a t, we have,

0 m/s - 10 m/s = a × 10 s; or a = (-10 m/s)/10s = - 1 m/s²

So the (negative) acceleration a on the body = - 1 m/s²

Distance travelled by the body before coming to rest can be obtained using one of the two relations:

v² - u² = 2 a s. Here v= 0 m/s, u = 10 m/s and a = -1 s². Substituting these values we have,

0² - 10² = 2 × (-1 ) × s; ==> s = 100/2 = 50 m

2. s = u t + ½ a t². Substituting we have s = 10 m/s× 10s + ½ (-1 m/s² ) × 10² s² = 100 m - 50 m = 50m

Both relation give the same result as it should

Answer 2

Answer:here you go

Explanation;

Hope it helps....