Consult Interactive Solution 2.22 before beginning this problem. A car is traveling along a straight road at a velocity of +38.8 m/s when its engine cuts out. For the next 2.01 seconds, the car slows down, and its average acceleration is . For the next 6.36 seconds, the car slows down further, and its average acceleration is . The velocity of the car at the end of the 8.37-second period is +26.6 m/s. The ratio of the average acceleration values is = 1.45. Find the velocity of the car at the end of the initial 2.01-second interval.
To find the velocity of the car at the end of the initial 2.01-second interval, we need to use the formula for average acceleration.
The formula for average acceleration is:
average acceleration = (change in velocity)/(time interval)
We are given the following information:
Initial velocity (v1) = +38.8 m/s
Time interval for the first stage (t1) = 2.01 seconds
Average acceleration for the first stage (a1) =
The formula can be rearranged to solve for the change in velocity:
change in velocity = average acceleration * time interval
Substituting the given values, we have:
change in velocity = * 2.01
Now we can subtract the change in velocity from the initial velocity to get the final velocity at the end of the initial 2.01-second interval:
final velocity = initial velocity - change in velocity
Substituting the values, we have:
final velocity = 38.8 - * 2.01
final velocity = 38.8 - (1.45 * 2.01)
final velocity = 38.8 - 2.9145
final velocity = 35.8855
The velocity of the car at the end of the initial 2.01-second interval is approximately +35.9 m/s.
To find the velocity of the car at the end of the initial 2.01-second interval, we can use the following formula:
vf = vi + at
Where:
vf is the final velocity
vi is the initial velocity
a is the average acceleration
t is the time
Given that the initial velocity vi is +38.8 m/s, and the time t is 2.01 seconds, we need to find the average acceleration a.
Let's use the formula for average acceleration:
a = (vf - vi) / t
Given that the final velocity vf is not provided, we can substitute it with the ratio of the average acceleration values (1.45) multiplied by the initial velocity vi:
vf = 1.45 * vi
Substituting this into the formula for average acceleration:
a = (1.45 * vi - vi) / t
Now, we can solve for a:
a = (1.45 * 38.8 - 38.8) / 2.01
Calculating this, we get:
a ≈ 34.292
Now, we can use the formula vf = vi + at to find the velocity of the car at the end of the initial 2.01-second interval:
vf = vi + at
vf = 38.8 + (34.292 * 2.01)
Calculating this, we get:
vf ≈ 38.8 + 69.067
vf ≈ 107.867
Therefore, the velocity of the car at the end of the initial 2.01-second interval is approximately +107.867 m/s.