Question

A body with an intial velocity of 3 m/s moves with an acceleration of 2 m/s, then the distance travelled in the 4th second is ​

Answer 1

Answer:

You can use the formula:

s=ut+1/2at^2

where s=distance travelled

u=initial velocity

a=acceleration

t=time

ATQ, u=3m/s,t=20s, a=2m/s^2, s=?

Now put the value in the formula:

s=ut+1/2at^2

s=3(4s)+ 1/2(2m/s^2)(4s)^2

s=12+1/2(3m/s^2)(16s^2)

s=12+1/2(24m)

s=24m

this is the distance travelled in 4 sec

Now calculate the distance in 3 sec than subtract the distance from 4 sec you will get the distance in 4th second

Distance in 3s

s1=3*2+1/2*3*3*3

s1=19.5m

therefore distance in 4th sec.is 24-19.5=4.5m