Question

test 1: unit 1
name:
total: /38

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 distance for part (a)

Use the formula \(s = \frac{(u + v)}{2}t\), where \(u = 12.0\space m/s\), \(v = 4.0\space m/s\), \(t = 3.0\space s\).

$$s=\frac{(12.0 + 4.0)}{2}\times3.0$$

$$s = 8.0\times3.0=24\space m$$

Step2: Calculate acceleration for part (b)

Use the formula \(a=\frac{v - u}{t}\), where \(u = 12.0\space m/s\), \(v = 4.0\space m/s\), \(t = 3.0\space s\).

$$a=\frac{4.0 - 12.0}{3.0}=\frac{- 8.0}{3.0}\approx - 2.67\space m/s^{2}$$

Step3: Calculate time to stop for part (c)

Use the formula \(t=\frac{v - u}{a}\), now \(v = 0\space m/s\), \(u = 4.0\space m/s\), \(a=\frac{- 8.0}{3.0}\space m/s^{2}\)

$$t=\frac{0 - 4.0}{\frac{- 8.0}{3.0}}=\frac{- 4.0\times3.0}{- 8.0}=1.5\space s$$

Step4: Calculate distance for part (d)

Use the formula \(s=vt\), where \(v = 4.0\space m/s\), \(t = 3.0\space s\)

$$s=4.0\times3.0 = 12\space m$$

Answer:

(a) \(24\space m\); (b) \(- 2.67\space m/s^{2}\); (c) \(1.5\space s\); (d) \(12\space m\)