as the shuttle bus comes to a sudden stop to avoid hitting a dog, it accelerates uniformly at -4.1 m/s to the 2nd power as it slows from 9.0m/s to 0.0 m/s. find the time interval of acccelaration for the bus
u=initial velocity
v=final velocity
a=acceleration
v=u+at
solve for t (watch out for sign of a)
To find the time interval of acceleration, we can use the equation for uniform acceleration:
vf = vi + at
Where:
- vf is the final velocity (0.0 m/s in this case)
- vi is the initial velocity (9.0 m/s in this case)
- a is the acceleration (-4.1 m/s^2 in this case)
- t is the time
Rearranging the equation, we get:
t = (vf - vi) / a
Let's substitute the given values:
t = (0.0 m/s - 9.0 m/s) / -4.1 m/s^2
Now we can calculate the time interval:
t = (-9.0 m/s) / (-4.1 m/s^2) = 2.19512 seconds (rounded to five decimal places)
Therefore, the time interval of acceleration for the bus is approximately 2.19512 seconds.
To find the time interval of acceleration for the bus, we can use the equation of motion involving acceleration, initial velocity, final velocity, and time.
The equation is:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the initial velocity (u) is 9.0 m/s, the final velocity (v) is 0.0 m/s, and the acceleration (a) is -4.1 m/s^2. We need to find the time interval (t).
Rearranging the equation, we have:
t = (v - u) / a
Substituting the given values, we get:
t = (0.0 - 9.0) / (-4.1)
Simplifying the calculation:
t = -9.0 / -4.1
t = 2.195
So, the time interval of acceleration for the bus is approximately 2.195 seconds.