Question

a body starting from rest moves with uniform acceleration travels 40 m in first 4 sec velocity after 16 sec will be​

Answer 1

Explanation:

▪️s = ut + ½ at^2 …. (1)

▪️v^2 = u^2 + 2as …. (2)

▪️v = u + at …. (3)

▪️s = (u + v)t/2 …. (4)

▪️where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

▪️In this case, we know u = 0, v = 40m/s, t = 10s and we want to find acceleration, a, so we use equation (3) v = u + at

▪️40 = 0 + 10a

▪️So a = 4m/s^2

▪️Now we have u = 0, t = 6s, a = 4m/s^2 and we want to find v, so we use equation (3) v = u + at

▪️v = 0 + 4(6) = 24m/s

▪️So the velocity at 6s is 24m/s

$\mathbb{hopes \: it \: help \: you}$

Answer 2

Answer:

️s = ut + ½ at^2 …. (1)

️v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

️s = (u + v)t/2 …. (4)

️where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, we know u = 0, v = 40m/s, t = 10s and we want to find acceleration, a, so we use equation (3) v = u + at

️40 = 0 + 10a

So a = 4m/s^2

️Now we have u = 0, t = 6s, a = 4m/s^2 and we want to find v, so we use equation (3) v = u + at

️v = 0 + 4(6) = 24m/s