Question

1. a cyclist is travelling with a speed of 12.0 m/s when she applies the brakes. after slowing for 3.0 s, her speed has been reduced to 4.0 m/s. (8 marks) (a) what distance does she travel during this time? (b) what is her acceleration? (c) if she continues braking, how much longer will it take her to stop? (d) if she continues to travel with her new speed, how far will she travel during the next 3.0 s?

Explanation:

Step1: Calculate acceleration

Use $a = \frac{v - u}{t}$ where $u=12.0\,\text{m/s}$, $v=4.0\,\text{m/s}$, $t=3.0\,\text{s}$
$a = \frac{4.0 - 12.0}{3.0} = -\frac{8.0}{3.0} \approx -2.67\,\text{m/s}^2$

Step2: Calculate distance during slowing

Use $s = \frac{(u + v)t}{2}$
$s = \frac{(12.0 + 4.0) \times 3.0}{2} = 24.0\,\text{m}$

Step3: Time to stop while braking

Use $t = \frac{0 - v}{a}$ where $v=4.0\,\text{m/s}$
$t = \frac{0 - 4.0}{-2.67} \approx 1.5\,\text{s}$

Step4: Distance at constant new speed

Use $s = v \times t$ where $v=4.0\,\text{m/s}$, $t=3.0\,\text{s}$
$s = 4.0 \times 3.0 = 12.0\,\text{m}$

Answer:

(a) $24.0\,\text{m}$
(b) $-2.67\,\text{m/s}^2$
(c) $1.5\,\text{s}$
(d) $12.0\,\text{m}$