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Chapter 2: Problem 81

A car moving at \(60.0 \mathrm{~km} / \mathrm{h}\) comes to a stop in \(t=4.00 \mathrm{~s}\) Assume uniform deceleration. a) How far does the car travel while stopping? b) What is its deceleration?

Short Answer

Expert verified
Question: A car is traveling at 60 km/h when the driver applies the brakes to come to a stop in 4 seconds. Calculate a) the distance the car travels while stopping, and b) the deceleration of the car. Answer: a) The car travels 33.4 meters while stopping. b) The deceleration of the car is -4.18 m/s².

Step by step solution

01

Convert the initial velocity to meters per second

We are given the initial velocity in km/h, but we'll need it in m/s to work with the equations of motion. To convert from km/h to m/s, we'll multiply by (1000 m/km) and divide by (3600 s/h): \(v_0 = 60.0 \frac{km}{h} \cdot \frac{1000 \thinspace m}{1 \thinspace km} \cdot \frac{1 \thinspace h}{3600 \thinspace s} = 16.7 \frac{m}{s}\)
02

Use the equations of motion to find the distance (a)

We'll use the equation \(v^2 = v_0^2 + 2ax\) to solve for the distance (x) the car travels while stopping. The final velocity (v) is 0 because the car comes to a stop: \(0 = (16.7 \thinspace m/s)^2 + 2a \cdot x\) Now, we'll solve for x: \(x = \frac{-(16.7 \thinspace m/s)^2}{2a}\) We still need to find the value of the deceleration (a) to calculate the distance, so let's proceed to the next step.
03

Use the equations of motion to find the deceleration (a)

We can use the equation \(v = v_0 + at\) to solve for the deceleration (a). We have the final velocity (v=0), the initial velocity (v₀=16.7 m/s), and the time (t=4.00 s): \(0 = 16.7 \frac{m}{s} + a(4.00 \thinspace s)\) Now we'll solve for a: \(a = -\frac{16.7 \frac{m}{s}}{4.00 \thinspace s} = -4.18 \frac{m}{s^2}\) Since the car is decelerating, the value of a is negative as expected.
04

Calculate the distance (a) using the found deceleration

Now that we have the deceleration (a), we can plug it back into the equation for the distance (x) we found in Step 2: \(x = \frac{-(16.7 \thinspace m/s)^2}{2(-4.18 \thinspace m/s^2)} = 33.4 \thinspace m\) So, the car travels 33.4 meters while stopping. In summary: a) The car travels 33.4 meters while stopping. b) The deceleration of the car is -4.18 m/s².

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