A car accelerates uniformly from +10.0 m/s to +50.0 m/s over a distance of 225 m. How long did it take to go that distance? Show all your work, including the equation used, given and unknown quantities, and any algebra required. Make sure your answer has the correct number of significant figures.
The average velocity was30m/s
time=distance/avbvelocity
To find the time taken by the car to travel the given distance, we can use one of the equations of motion:
\[ v_f = v_i + at \]
Where:
\( v_f \) = final velocity = 50.0 m/s
\( v_i \) = initial velocity = 10.0 m/s
\( a \) = acceleration (which is constant)
\( t \) = time taken
First, let's rearrange the equation to solve for \( t \):
\[ t = \frac{{v_f - v_i}}{a} \]
Now we need to determine the value of \( a \). Since the car is said to accelerate uniformly, we can use the formula:
\[ a = \frac{{\Delta v}}{\Delta t} \]
Where:
\( \Delta v \) = change in velocity = \( v_f - v_i \)
\( \Delta t \) = change in time
Rearranging this equation, we get:
\[ \Delta t = \frac{{\Delta v}}{a} \]
Given that the car starts from \( v_i = 10.0 \, \text{m/s} \) and reaches \( v_f = 50.0 \, \text{m/s} \), the change in velocity is \( \Delta v = v_f - v_i = 50.0 \, \text{m/s} - 10.0 \, \text{m/s} = 40.0 \, \text{m/s} \).
Substituting this value into the equation for \( \Delta t \), we get:
\[ \Delta t = \frac{{\Delta v}}{a} = \frac{{40.0 \, \text{m/s}}}{{a}} \]
Now, let's substitute \( \Delta t \) and the given distance \( \Delta d = 225 \, \text{m} \) into another equation:
\[ \Delta d = \frac{1}{2}(v_i + v_f)t \]
Substituting the values:
\[ 225 \, \text{m} = 0.5(10.0 \, \text{m/s} + 50.0 \, \text{m/s})t \]
\[ 225 \, \text{m} = 0.5(60.0 \, \text{m/s})t \]
\[ 225 \, \text{m} = 30.0 \, \text{m/s} \cdot t \]
Simplifying:
\[ t = \frac{{225 \, \text{m}}}{{30.0 \, \text{m/s}}} = 7.50 \, \text{s} \]
Therefore, the car took 7.50 seconds to travel the given distance of 225 meters.