## College Physics (4th Edition)

At $t=8.00~s$, the magnitude of the velocity is 44.7 m/s
We can find the south component of velocity at $t = 8.00~s$: $v_f = v_0+at$ $v_f = 0+(2.50~m/s^2)(8.00~s)$ $v_f = 20.0~m/s$ The east component of the velocity vector will continue to be 40.0 m/s as the south component of the velocity vector increases. We can find the magnitude of the velocity at $t = 8.00~s$: $\sqrt{(20.0~m/s)^2+(40.0~m/s)^2} = 44.7~m/s$ At $t=8.00~s$, the magnitude of the velocity is 44.7 m/s