Question

1. if a vehicles speed doubles, e.g., from 20 mph to 40 mph, the distance needed to stop the vehicle increases by times.

a. 2
b. 3
c. 4
d. 8

Explanation:

Step1: Use the kinematic formula \(v^{2}=u^{2}+2as\)

Assume the deceleration \(a\) is constant. Let the initial speed be \(u\) and final speed \(v = 0\). Then \(s=\frac{-u^{2}}{2a}\) (negative sign for deceleration, magnitude considered).

Step2: Analyze when speed doubles

If initial speed is \(u_1\) and distance \(s_1=\frac{u_1^{2}}{2|a|}\), when speed \(u_2 = 2u_1\), then \(s_2=\frac{(2u_1)^{2}}{2|a|}=\frac{4u_1^{2}}{2|a|}\)

Step3: Find the ratio

\(\frac{s_2}{s_1}=\frac{\frac{4u_1^{2}}{2|a|}}{\frac{u_1^{2}}{2|a|}} = 4\)

Answer:

C. 4