A car is traveling at 7.0 m/s when the driver applies the brakes. The car moves 1.5 m before it comes to a complete stop. It the car had been moving at 14 m/s, how far would it have continued to move after the brakes were applied ? Assume the braking force is constant .

Options

A. 1.5m
B. 4.5m
C. 7.5m
D. 6.0m

Answer

To solve the problem, we will use the equations of motion to find the distance the car would travel after the brakes are applied when it is initially moving at 14 m/s.

### Step 1: Determine the acceleration when the car is traveling at 7 m/s.

We know:
- Initial velocity (u) = 7 m/s
- Final velocity (v) = 0 m/s (the car comes to a stop)
- Distance (s) = 1.5 m

Using the equation of motion:
\[ v^2 = u^2 + 2as \]

Substituting the known values:
\[ 0 = (7)^2 + 2a(1.5) \]
\[ 0 = 49 + 3a \]

Rearranging to find acceleration (a):
\[ 3a = -49 \]
\[ a = -\frac{49}{3} \, \text{m/s}^2 \]

### Step 2: Calculate the distance traveled when the car is traveling at 14 m/s.

Now, we need to find the distance (s') the car would travel when its initial velocity (u') is 14 m/s, using the same acceleration (a) calculated above.

Using the same equation of motion:
\[ v^2 = u'^2 + 2as' \]

Here:
- Initial velocity (u') = 14 m/s
- Final velocity (v) = 0 m/s
- Acceleration (a) = -\frac{49}{3} m/s²

Substituting the known values:
\[ 0 = (14)^2 + 2\left(-\frac{49}{3}\right)s' \]
\[ 0 = 196 - \frac{98}{3}s' \]

Rearranging to solve for s':
\[ \frac{98}{3}s' = 196 \]
\[ s' = \frac{196 \cdot 3}{98} \]
\[ s' = 6 \, \text{m} \]

### Final Answer:
The car would continue to move **6 meters** after the brakes are applied when it was initially traveling at 14 m/s.

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