## Jun 9, 2018

#### Explanation:

The velocity is constant at ${t}_{2} = 30 s$

$\frac{\mathrm{ds}}{\mathrm{dt}} = 60 k m {h}^{-} 1$

Therefore,

The acceleration is

$a = \frac{{d}^{2} s}{\mathrm{dt}} ^ 2 = \frac{d}{\mathrm{dt}} \left(\frac{\mathrm{ds}}{\mathrm{dt}}\right) = 0 m {s}^{-} 2$

Jun 9, 2018

The instantaneous acceleration at ${t}_{2} = 30$ sec. is
${a}_{2} = - \frac{10}{6} \frac{m}{s} ^ 2 \approx - 1.67 \frac{m}{s} ^ 2$

#### Explanation:

According to the figure we have a constant speed until 30 sec, at which point the movement descelerates, so that the speed changes constantly from 60 km/h to 0 km/h in 10 sec. (from 30 to 40 sec.).

Acceleration is speed change per second, i.e. $\frac{\Delta v}{\Delta t}$

We, therefore need to convert the speed from $\frac{k m}{h}$ to $\frac{m}{s}$

$60 \frac{k m}{h} = \frac{60000}{3600} \frac{m}{s} = \frac{100}{6} \frac{m}{s}$

(60 km = 60 000 m, and there are 3600 sec. in 1 hour)

Therefore $a = \frac{\Delta v}{\Delta t}$
$= \frac{0 - \frac{100}{6}}{10} \frac{m}{s} ^ 2$
$= - \frac{10}{6} \frac{m}{s} ^ 2 = - 1.67 \frac{m}{s} ^ 2$

The figure shows that the acceleration is constant from 30 to 40 sec. The instantaneous acceleration at ${t}_{2} = 30 s$, therefore, is
${a}_{2} = - \frac{10}{6} \frac{m}{s} ^ 2 \approx - 1.67 \frac{m}{s} ^ 2$