Question

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

Answer 1

Answer:

Explanation:

=> Here, a force F is applied to the initially, stationary cart. The variation of force with time is parabolic as shown in the figure. Thus, The equation of parabola is  

x²=4ay  

But, t²=4aF  

=> When the time t is 5s and force F is 50 N as given in the graph,  

5²=4 *a *50  

25 = 200 * a

a= 25/200

a = 1/8  

∴ t² = 4 * 1/8 F

t² = F/2

∴ F=2*t²  

Acceleration = F/m = 2t²/m = 2t²/10 = t²/5  

∴dv/dt=t²/5 or $\int\limits^v_0 {dv}=\int\limits^5_0 {(t^2/5)}dt$

∴v=1/5[t³/3]^5_0 = 125/15

=8.33ms⁻¹

Thus, the speed of cart at given time is 8.33 ms⁻¹