A car accelerates uniformly from rest to a speed of 64.5 km/h (17.9 m/s) in 10 s. Find the distance the car travels during this time.
easy way:
average speed = 17.9/2
distance = 179/2 meters
Hard way:
v = Vi + a t
17.9 = 0 + a(10)
a = 1.79
x = Xi + Vi t + (1/2) a t^2
x-Xi = distance = 0 (10 + (1/2)(1.79) (100)
=179/2
To find the distance the car travels during this time, we can use the equation for distance traveled when an object accelerates uniformly:
\[Distance = \frac{{1}}{{2}} \times \text{{acceleration}} \times \text{{time}}^2\]
In this case, we know the time is 10 seconds, but we need to find the acceleration first.
To find the acceleration, we can use the equation:
\[acceleration = \frac{{\text{{final velocity}} - \text{{initial velocity}}}}{\text{{time}}}\]
Given that the initial velocity is 0 (as the car starts from rest) and the final velocity is 17.9 m/s, we can substitute these values into the equation:
\[acceleration = \frac{{17.9 \, \text{m/s} - 0 \, \text{m/s}}}{10 \, \text{s}}\]
Simplifying this, we get:
\[acceleration = 1.79 \, \text{m/s}^2\]
Now we can substitute the values of acceleration (1.79 m/s^2) and time (10 seconds) into the distance equation:
\[Distance = \frac{{1}}{{2}} \times 1.79 \, \text{m/s}^2 \times (10 \, \text{s})^2\]
Simplifying this, we get:
\[Distance = \frac{{1}}{{2}} \times 1.79 \, \text{m/s}^2 \times 100 \, \text{s}^2\]
\[Distance = 89.5 \, \text{m}\]
Therefore, the car travels a distance of 89.5 meters during this time.