Provided a racing car does not lose traction, the time taken by it to race from rest through a distance depends mainly on engine's power P. Derive the time t in terms of S and P, assuming that power is constant.

Explanation

Hint : `P=mav=mv(dv)/(dt)" "(a=(dV)/(dt))`
`rArr" "int_(0)^(v)vdv=(p)/(m)int_(0)^(t)dt`
Find v, then `v=(dx)/(dt)`
`int_(0)^(x)dx=int_(0)^(t)vdt`
Find x.
The power of an engine is the rete at which energy is supplied as given by
`P=mav=mv(dv)/(dt)`
`rArr" "int_(0)^(v)vdv=(P)/(m)int_(0)^(1)dt`
`rArr" "(v^(2))/(2)=(P)/(m)t`
`rArr" "v=sqrt((2P)/(m))t^(1//2)`
`rArr" "(dx)/(dt)=sqrt((2P)/(m))t^(1//2)`
`rArr" Force on the truck when it moves down the hill"`
`F.=mg sin theta-mu_(k)mg cos theta`
`=mg(sin theta-mu_(k))("taking "cos theta=1" as angle is very small")`
`=10^(5)((1)/(50)-0.2)`
`=10^(5)((-9)/(50))=-18,000N`
Negative sign signifies direction
Power = F..v.
`22xx10^(4)=18000v`
`v=(220)/(18)=12.2ms^(-1)`
`=12.2xx(18)/(5)=43.99`
`=44kmh^(-1)int_(0)^(x)dx=sqrt((2P)/(m))int_(0)^(t)t^(1//2)dt`
`rArr" "x=sqrt((8)/(9m))t^(3//2)`