Question

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

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

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): \(v_0 = 60.0 \frac{km}{h} \cdot \frac{1000 \thinspace m}{1 \thinspace km} \cdot \frac{1 \thinspace h}{3600 \thinspace s} = 16.7 \frac{m}{s}\)

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.

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.

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.

Key concepts

Equations of Motion

Equations of motion are essential for describing the movement of objects under uniform acceleration. In the given problem, the key equations help us to determine how far the car travels and its deceleration. These equations assume that acceleration is constant.
One of the primary equations used in this context is:

• \[v = v_0 + at\]

Here, \(v\) is the final velocity, \(v_0\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. This equation is used to find acceleration when time and velocities are known.
Another essential equation is:

• \[v^2 = v_0^2 + 2ax\]

This equation helps calculate the distance (\(x\)) once you know the initial and final velocities and the acceleration. In the exercise, these equations allowed us to find how far the car traveled and its deceleration.

Initial Velocity Conversion

Before applying the equations of motion, the initial velocity must be in an appropriate unit. In physics, especially when dealing with the equations of motion, converting velocities to meters per second (m/s) is common.
The original velocity of the car is given as 60.0 km/h. To convert this to m/s:

• Multiply by \(1000\) to convert kilometers to meters.
• Divide by \(3600\) to convert hours to seconds.

The resulting conversion is: \[v_0 = 60.0 \frac{km}{h} \cdot \frac{1000 \, m}{1 \, km} \cdot \frac{1 \, h}{3600 \, s} = 16.7 \frac{m}{s}\] This initial step is crucial because all subsequent calculations depend on having the correct unit of measurement for velocity.

Calculating Distance and Acceleration

To find how far the car travels and its deceleration, we use the equations of motion with the information given. Since the car comes to a stop, its final velocity is zero. The initial velocity and time are already provided.
First, we determined the deceleration (\(a\)) using the formula \(v = v_0 + at\). By substituting \(v = 0\), \(v_0 = 16.7 \ m/s\), and \(t = 4.00 \ s\):

• \[0 = 16.7 \frac{m}{s} + a \times 4.00 \ s\]

Solving gives: \[a = -\frac{16.7 \frac{m}{s}}{4.00 \ s} = -4.18 \frac{m}{s^2}\] The negative sign indicates deceleration.
Then, using the equation \(v^2 = v_0^2 + 2ax\) and substituting the known values:

• \[0 = (16.7 \ m/s)^2 + 2 \times (-4.18 \ m/s^2) \times x\]

We find: \[x = \frac{(16.7 \ m/s)^2}{2 \times 4.18 \ m/s^2} = 33.4 \ m\] This calculation shows that the car travels 33.4 meters while stopping, with a deceleration of \(-4.18 \ m/s^2\).