Question

a car travels at 20 m/s and approaches a red light. it slows to a stop in 3s.
a) what was the acceleration of the car as it slowed?
b) how far did the car move while it slowed down?
c) if when the light turns green, the car starts to accelerate at 2 m/s^2, how long will it take to get back up to 20m/s?

Explanation:

Step1: Calculate acceleration in part a

Use the formula \(a=\frac{v - u}{t}\), where \(u = 20\ m/s\), \(v=0\ m/s\), \(t = 3\ s\).
\(a=\frac{0 - 20}{3}=-\frac{20}{3}\approx - 6.67\ m/s^{2}\)

Step2: Calculate distance in part b

Use the formula \(s=ut+\frac{1}{2}at^{2}\). Substitute \(u = 20\ m/s\), \(a=-\frac{20}{3}\ m/s^{2}\), \(t = 3\ s\)

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Or use \(s=\frac{(u + v)}{2}\times t=\frac{(20+0)}{2}\times3 = 30\ m\)

Step3: Calculate time in part c

Use the formula \(v=u+at\). Here \(u = 0\ m/s\), \(v = 20\ m/s\), \(a = 2\ m/s^{2}\)
\(t=\frac{v - u}{a}=\frac{20-0}{2}=10\ s\)

Answer:

a) \(-6.67\ m/s^{2}\)
b) \(30\ m\)
c) \(10\ s\)