Question

A 1.70 kg box is moving to the right with speed 9.00 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. What distance does the box move from its position at t = 0 before its speed is reduced to zero? If the force continues to be applied, what is the velocity of the box at 5.00 s.

Answer 1

If F(t) = -6N/s² * t², then
a(t) = -(6N /1.7kg·s²) * t², and then
v(t) = ∫ a(t) dt = -(2N /1.7 kg·s²)t³ + C

Since v(0) = 9.0 m/s, C = 9 m/s, and
v(t) = 9.0 m/s - (2N /1.7kg·s²)t³

Then s(t) = ∫ v(t) dt = 9m/s*t - (0.294 Nkg·s²)t^4 + C
and we can assume that C = 0 m here.

A)
v(t) = 0 = 9.0 m/s - (1.176 Nkg·s²)t³
works out to t = 1.97 s
then s(1.97) = 9m/s*1.97 - (0.294 Nkg·s²)*1.97^4 = 13.3 m
Distance box move from its position before it's speed reduced to zero, = 13.3 m


B)
v(t) = 9.0 m/s - (2N /1.7kg·s²)t³
v(5) = 9 - (2 /1.7)*5^3 m/s
v(5) = -138.05 m/s
velocity of the box at 5.00 s, = -138.05 m/s (-ve sign means it moves towards left direction.)