A bike accelerates from 0.0 m/s to 4.1 m/s in 6 s. What distance does it travel?
thanks for the help
Use
v²-u² = 2aS
where
S=distance travelled,
u=initial velocity
v=final velocity
Solve for S
To find the distance the bike travels, we can use the formula:
distance = initial velocity * time + (1/2) * acceleration * time^2
In this case, the initial velocity is 0.0 m/s, the final velocity is 4.1 m/s, and the time taken is 6 seconds. However, since we are not given the acceleration directly, we need to calculate it.
We can use the equation:
final velocity = initial velocity + (acceleration * time)
Using the given values, we can rearrange the equation to solve for acceleration:
acceleration = (final velocity - initial velocity) / time
acceleration = (4.1 m/s - 0.0 m/s) / 6 s
acceleration = 4.1 m/s / 6 s
acceleration ≈ 0.683 m/s^2
Now that we have the acceleration, we can substitute it back into the distance formula to calculate the distance traveled:
distance = initial velocity * time + (1/2) * acceleration * time^2
distance = 0.0 m/s * 6 s + (1/2) * 0.683 m/s^2 * (6 s)^2
distance = 0.0 m + (1/2) * 0.683 m/s^2 * 36 s^2
distance = 0.0 m + 0.5 * 0.683 m/s^2 * 36 s^2
distance = 0.0 m + 0.5 * 0.683 m * 36 s
distance = 0.0 m + 0.3415 m * 36 s
distance = 0.0 m + 12.294 m
distance ≈ 12.294 m
Therefore, the bike travels approximately 12.294 meters.