Question

A BODY STARTS ITS MOTION WITH AS SPEED OF 10MS-¹ AND ACCELERATE FOR 10S WITH 10MS-² WHAT WILL BE THE DISTANCE AFTER 10 SECONDS​

Answer 1

Explanation:

u = 10m/s

a = 10m/s^2

t = 10s

s= ut+1/2at^2 (2nd Equation of Motion)

s= 10(10) +1/2×10×10^2

s= 100 + 1/2×1000

s= 100+500

s= 600m

THE DISTANCE AFTER 10 SECONDS IS 600M

Answer 2

Correct question:-

A body starts it's motion with as a speed of 10m/s and accelerate for 10s with 10m/s².What will be the distance after 10 seconds.

Given:-

•Initial velocity of the body(u)=10m/s

•Acceleration of the body(a)=10m/s²

•Time taken(t)=10s

To find:-

•Distance covered by the body in 10s.

Solution:-

By,using the 2nd equation of motion,we get:-

=>s=ut+1/2at²

=>s=10×10+1/2×10×(10)²

=>s=100+500

=>s=600m

Thus,distance covered by the body in 10s is 600m.

Some Extra information:-

The 3 equations of motion are:-

•v=u+at

•s=ut+1/2at²

•v²-u²=2as

Note:-The above 3 equations are only valid when a body is moving with uniform acceleration.