Question

A body started from rest and attained a velocityof 15 m/s in 5 sec . Find the distance covered by body for attaing this velocity

Answer 1

Answer:

Since a body starts from rest so

Initial velocity (u)=0 m/s

Final velocity (v)= 10m/s

Time taken to acquire velocity (T)=2 seconds

Acceleration(A)=change in velocity/time taken

V-u/T. =10–0/2 =5m/s^2

Hence accelration is 5m/s^2.

Explanation:

Hope it helps u

Answer 2

Answer:-

$\red{\bigstar}$ The distance covered by the body for attaining this velocity will be

$\large\leadsto\boxed{\sf{37.5 \: m}}$

• Given:-

Initial velocity [u] = 0 $\\\sf{[As \: the \: body \: started \: from \: rest.]}$

Final velocity [v] = 15 m/s

Time [t] = 5 sec.

• To Find:-

Distance covered by the body = ?

• Solution:-

• Using the 1st equation of motion:-

$\pink{\bigstar}$$\sf{v = u + at}$

→ 15 = 0 + a × 5

→ 15 = 5a

→ a = $\dfrac{15}{5}$

→ a = 3 m/s²

Now,

In order to find the distance covered,

• Using the 2nd equation of motion:-

$\pink{\bigstar}$$\sf{s = ut + \dfrac{1}{2}a t^{2}}$

→ s = $\sf{0 \times 5 + \dfrac{1}{2} \times 3 \times (5)^2}$

→ s = $\sf{0 + 5 + \dfrac{1}{2}\times 3 \times 25}$

→ s = $\sf{0 + \dfrac{75}{2}}$

→ s = $\sf{\dfrac{75}{2}}$

→ $\boxed{\sf s = 37.5}$

Therefore, the distance covered by the body for attaining this velocity will be 37.5 m