## College Physics (4th Edition)

(a) The change in speed is 12.0 m/s (b) $v = 33.0~m/s$
(a) We can use the information in the question to write the linear equation for the speed: $v = (6.0~m/s^2)~t+3.0~m/s$ We can find the change in speed in the time interval between 4.0 seconds and 6.0 seconds: $\frac{\Delta v}{\Delta t} = m$ $\Delta v = m~\Delta t$ $\Delta v = (6.0~m/s^2)(6.0~s-4.0~s)$ $\Delta v = 12.0~m/s$ The change in speed is 12.0 m/s (b) We can find the speed when the elapsed time is 5.0 seconds: $v = (6.0~m/s^2)~t+3.0~m/s$ $v = (6.0~m/s^2)(5.0~s)+3.0~m/s$ $v = 33.0~m/s$