A force F is applied to the initially, stationary cart. The variation of force with time is shown in the figure. The speed of cart at t = 5 s is

Options

A. 10 m/s
B. 8.33 m/s
C. 2 m/s
D. zero

Answer

B

Explanation

The equation of parabola is
`x^(2)=4ay`
Here, `t^(2)=4aF`
When t = 5 s, F = 50 N (see graph)
`therefore" "5^(2)=4axx50`
`therefore " "a=(25)/(200)=(1)/(8)`
`therefore t^(2) = 4 xx(1)/(8)F=(F)/(2)implies F=2t^(2)`
Acceleration `=(F)/(m)=(2t^(2))/(m)=(2t^(2))/(10)=(t^(2))/(5)`
`therefore(dv)/(dt)=(t^(2))/(5)" or "int_(0)^(v)dv=int_(0)^(5)(t^(2))/(5)dt`
`therefore v = (1)/(5)[(t^(3))/(3)]_(0)^(5)=(125)/(15)=8.33 m//s`