## College Physics (4th Edition)

At $t=0.8~s$, the magnitude of the velocity will be 100 m/s
The east component of the velocity vector will continue to be 60 m/s as the north component of the velocity vector increases. Let $v_n$ be the north component of the velocity vector when the magnitude of the velocity is 100 m/s. We can find $v_n$: $\sqrt{v_n^2+(60~m/s)^2} = 100~m/s$ $v_n^2+(60~m/s)^2 = (100~m/s)^2$ $v_n^2 = (100~m/s)^2 - (60~m/s)^2$ $v_n = \sqrt{(100~m/s)^2 - (60~m/s)^2}$ $v_n = 80~m/s$ We can find the time when the north component of velocity is 80 m/s: $v_f = v_0+at$ $t = \frac{v_f-v_0}{a}$ $t = \frac{80~m/s-0}{100~m/s^2}$ $t = 0.8~s$ At $t=0.8~s$, the magnitude of the velocity will be 100 m/s