Answer
(a)
(i) $1.7 ~m/s^2$
(ii) $-1.7 ~m/s^2$
(iii) $0 ~m/s^2$
(iv) $0 ~m/s^2$
(b)
(i) $0 ~m/s^2$
(ii) $-1.7 ~m/s^2$
Work Step by Step
$v = (60 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s}) = 16.7 ~m/s$
(a) The average acceleration is $\frac{\Delta ~v}{\Delta ~t}$
(i) $\frac{\Delta ~v}{\Delta ~t} = \frac{16.7 ~m/s}{10 ~s} = 1.7 ~m/s^2$
(ii) $\frac{\Delta ~v}{\Delta ~t} = \frac{-16.7 ~m/s}{10 ~s} = -1.7 ~m/s^2$
(iii) $\frac{\Delta ~v}{\Delta ~t} = \frac{0 ~m/s}{20 ~s} = 0 ~m/s^2$
(iv) $\frac{\Delta ~v}{\Delta ~t} = \frac{0 ~m/s}{40 ~s} = 0 ~m/s^2$
(b) The instantaneous acceleration is the slope of the velocity versus time graph.
(i) At t = 20 s, the slope of the velocity versus time graph is zero. Therefore the instantaneous acceleration is $0 ~m/s^2$
(ii) At t = 35 s, the slope of the velocity versus time graph is $\frac{-16.7 ~m/s}{10 ~s} = -1.7 ~m/s^2$. Therefore the instantaneous acceleration is $-1.7 ~m/s^2$
