A bus slows down uniformly from 65.0 km/h
to 0 km/h in 20.0 s.
How far does it travel before stopping?
Answer in units of m
a= (Vf-Vi)/time
change Vi to m/s, solve for a.
d= vi*t+1/2 a t^2 solve for d
920
To find the distance the bus travels before stopping, we can use the formula for distance covered during uniformly accelerated motion:
distance = initial velocity * time + 0.5 * acceleration * time^2
In this case, the bus is slowing down uniformly, so its acceleration is negative. The acceleration can be calculated using the formula:
acceleration = (final velocity - initial velocity) / time
Given:
Initial velocity (u) = 65.0 km/h
Final velocity (v) = 0 km/h
Time (t) = 20.0 s
First, we need to convert the initial velocity from km/h to m/s since the formula requires SI units. We know that 1 km/h = 1000 m/3600 s. So,
initial velocity = 65.0 km/h * (1000 m/3600 s) = 18.056 m/s
Next, we calculate the acceleration:
acceleration = (final velocity - initial velocity) / time
= (0 m/s - 18.056 m/s) / 20.0 s
= -0.9028 m/s^2 (Note: the negative sign indicates deceleration)
Now, we can plug in the values into the distance formula:
distance = initial velocity * time + 0.5 * acceleration * time^2
= 18.056 m/s * 20.0 s + 0.5 * (-0.9028 m/s^2) * (20.0 s)^2
= 361.12 m - 180.56 m
= 180.56 m
Therefore, the bus travels a distance of 180.56 meters before stopping.