Question

A car start from rest acquires a velocity of 10 m/s in 5 second calculate (the acceleration) b (the distance covered by the the body in 5 second​

Answer 1

Answer:nitial velocity = 0 m/s

final velocity = 10 m/s

time = 5 seconds

(a) need to find ==> acceleration

given ==> initial velocity, final velocity and time

procedure ==> a = (v-u)/t

a = (10-0)/5

a = 10/5 = 2 m/s^2

therefore, the acceleration of the body is 2 m/s^2

(b) need to find ==> distance

given ==> acceleration, time, initial velocity and final velocity

procedure ==> s = ut + 1/2 at^2

s = 0*5 + 1/2 * 2 * 5 * 5

s = 0 + 5*5

s = 0 + 25

s = 25 m

therefore, the distance covered by the body is 25 metres

Explanation:

Answer 2

Answer:

initial velocity : 0m/s

time : 5 sec

final velocity : 10m/s

a) to find -- acceleration

We know that ;

v= u + a t

-- 10 = 0+ a * 5

-- a = 10/ 5

therefore , acceleration is 2 m/s^2

b) to find -- distance travelled in 5 second

We know that ;

s= u t + 1/2 a t^2

-- s= 0*5 + 1/2 * 2 * 5^2

-- s= 0 + 10

therefore , distance travelled in 5 sec is 10 m/s^2.

hope it will help you