A particle starts with an initial velocity 2.5 m/s along the positive x direction and it accelerates uniformly at the rate `0.50 m/s^2`. A. Find the distance travelled by it in the first two seconds. b.How much time does it take to reach the velocity `7.5 m/s` ? c. How much distance will it cover in reaching the velocity 7.5 m/s?

Explanation

a. We have,
`x=ut+1/2 at^2`
`=(2.5 m/s)(2)+1/2(0.50 m/s^2)(2s)^2`
`=5.0 m+1.0m=6.0m`.
Since the particle does not turn back it is also the distnce travelled.
b. We have ltbr. `v=u+at`
`or 7.5 m/s=2.5 m/s+(0.50m/s^2)t`
or `t= (7.5m/s-2.5m/s)/(0.50 m/s^2)=10s`
c. We have,
`v^2=u^2+2ax`
`or (7.5 m/s)^2=(2.5m/s)^2+2(0.50 m/s^2)x`
or, `x= ((7.5 m/s)^2-(2.5 m/s)^2)/(2x0.50m/s)^2=50m`.