Question

A car moving at \(60 \mathrm{~km} / \mathrm{h}\) comes to a stop in \(4.0 \mathrm{~s}\). What was its average deceleration? a) \(2.4 \mathrm{~m} / \mathrm{s}^{2}\) b) \(15 \mathrm{~m} / \mathrm{s}^{2}\) c) \(4.2 \mathrm{~m} / \mathrm{s}^{2}\) d) \(41 \mathrm{~m} / \mathrm{s}^{2}\)

Short answer

A) 2.50 m/s^2 B) 3.17 m/s^2 C) 4.17 m/s^2 D) 5.17 m/s^2 Answer: C) 4.17 m/s^2

Step-by-step solution

Convert initial velocity from km/h to m/s

First, we need to convert the initial velocity from kilometers per hour to meters per second. We can do this using the conversion factor 1000 meters per kilometer and 3600 seconds per hour. Multiply the initial velocity by 1000/3600: \(60 \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} \times \frac{1 \mathrm{~h}}{3600 \mathrm{~s}} = 16.67 \frac{\mathrm{m}}{\mathrm{s}}\)

Apply the formula for acceleration

Now that we have the initial velocity in meters per second, we can use the formula for acceleration: \(a = \frac{v_f - v_i}{t}\) where \(a\) is the acceleration, \(v_f\) is the final velocity, \(v_i\) is the initial velocity, and \(t\) is the time it takes for the change in velocity to occur.

Substitute the given values and solve for average deceleration

Plug in the given values: \(v_f = 0 \frac{\mathrm{m}}{\mathrm{s}}\), \(v_i = 16.67 \frac{\mathrm{m}}{\mathrm{s}}\), and \(t = 4.0 \mathrm{~s}\): \(a = \frac{0 \mathrm{~m/s} - 16.67 \mathrm{~m/s}}{4.0 \mathrm{~s}}\) \(a = -4.17 \frac{\mathrm{m}}{\mathrm{s}^{2}}\) The average deceleration is \(-4.17 \frac{\mathrm{m}}{\mathrm{s}^{2}}\). Since deceleration is negative acceleration, and we are looking for the magnitude of deceleration, we can drop the negative sign and obtain the answer: \(4.17 \frac{\mathrm{m}}{\mathrm{s}^{2}}\) The correct answer is closest to option C).