## College Physics (4th Edition)

The elevator's speed at $~t = 4.00~s~$ is $~2.27~m/s$
We can find the downward acceleration of the elevator: $\sum F = ma$ $mg-F_T = ma$ $a = \frac{mg-F_T}{m}$ $a = \frac{(832~kg)(9.80~m/s^2)-7730~N}{832~kg}$ $a = 0.51~m/s^2$ We can find the velocity at $t= 0$: $\Delta y = v_0~t+\frac{1}{2}at^2$ $v_0 = \frac{\Delta y-\frac{1}{2}at^2}{t}$ $v_0 = \frac{5.00~m-(\frac{1}{2})(0.51~m/s^2)(4.00~s)^2}{4.00~s}$ $v_0 = 0.23~m/s$ We can find the velocity at $t = 4.00~s$: $v_f = v_0+at$ $v_f = 0.23~m/s+(0.51~m/s^2)(4.00~s)$ $v_f = 2.27~m/s$ The elevator's speed at $~t = 4.00~s~$ is $~2.27~m/s$