A 1.90kg box is moving to the right with speed 9.50 m/s on a horizontal, frictio

Question

A 1.90kg box is moving to the right with speed 9.50 m/s on a horizontal, frictionless surface. At t = 0 a horizontal force is applied to the box. The force is directed to the left and has magnitude F(t) = (6.00 N/s^2) t^2 A) What distance does the box move from its position at before its speed is reduced to zero? Express your answer with the appropriate units.? x= ? B) If the force continues to be applied, what is the speed of the box at 3.50? Express your answer with the appropriate units.? v = ?

Explanation / Answer

F(t) = (6.00 N/s^2) t^2

-ma = (6.00 N/s^2) t^2

so , a =- 3.157t^2

a = dv/dt =- 3.157t^2

by integration, v= -1.05t^3 + c

at t = 0 , v=9.5

so c=9.5

v =- 1.05t^3 +9.5

when v = 0, t = 2.08 seconds.

v=dx/dt =- 1.05t^3 +9.5

by integration, x = -0.26t^4 + 9.5 t + c

at t=0 ,x =0 so c=0

so at t= 2.08 seconds, x = 14.89 m ANSWER

remaining part in comments.

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