If a car has a constant acceleration of 4 m/s², starting from rest, how far has it travelled by the time it reaches the speed of 40 m/s?
final velocity=at
t= 40/4 = 10 sec
distance= 1/2 a t^2=1/2 * 4*100
To find the distance traveled by the car, we can use the equation:
distance = (final velocity² - initial velocity²) / (2 * acceleration)
Here, the initial velocity is 0 m/s (starting from rest), the final velocity is 40 m/s, and the acceleration is 4 m/s².
Plugging in the values:
distance = (40² - 0²) / (2 * 4)
= (1600 - 0) / 8
= 1600 / 8
= 200 meters
Therefore, by the time the car reaches a speed of 40 m/s, it will have traveled a distance of 200 meters.
To find the distance traveled by the car, we can use the equation of motion:
\(v^2 = u^2 + 2as\)
Where:
- \(v\) is the final velocity (40 m/s)
- \(u\) is the initial velocity (0 m/s, since it starts from rest)
- \(a\) is the acceleration (4 m/s²)
- \(s\) is the distance traveled
Rearranging the equation to solve for \(s\):
\(s = \frac{(v^2 - u^2)}{2a}\)
Plugging in the given values:
\(s = \frac{(40^2 - 0^2)}{2 \cdot 4}\)
Now, let's calculate:
\(s = \frac{(1600 - 0)}{8}\)
\(s = \frac{1600}{8}\)
\(s = 200\)
Therefore, the car has traveled a distance of 200 meters by the time it reaches a speed of 40 m/s.