1) A car moving with an initial velocity of 29.6 m/s decreases to a velocity of 17.9 m/s over a time
interval of 4.08 seconds.
(a.) What is the average acceleration of the car during this time?
(b.) Assuming the acceleration is constant, how far did the car travel during this time?
change in velocity/ change in time = (17.9-29.6) / 4.08
= -2.67 m/s^2
d = Vi t - (1/2) 2.67 * t^2
= 29.6(4.08) - 1.33 (4.08^2)
= 121 - 22.1
= 98.9 m
To find the average acceleration of the car during the given time interval, we can use the formula:
Average acceleration = (final velocity - initial velocity) / time
(a.) Using the given values:
Initial velocity (u) = 29.6 m/s
Final velocity (v) = 17.9 m/s
Time (t) = 4.08 seconds
Average acceleration = (17.9 m/s - 29.6 m/s) / 4.08 seconds
Average acceleration = -11.7 m/s / 4.08 seconds
Average acceleration ≈ -2.87 m/s²
So, the average acceleration of the car during this time is approximately -2.87 m/s².
(b.) To find the distance traveled by the car during this time, assuming constant acceleration, we can use the kinematic equation:
Distance = (initial velocity * time) + (0.5 * acceleration * time²)
Using the given values:
Initial velocity (u) = 29.6 m/s
Time (t) = 4.08 seconds
Average acceleration = -2.87 m/s²
Distance = (29.6 m/s * 4.08 seconds) + (0.5 * -2.87 m/s² * (4.08 seconds)²
Distance = 120.768 m + (-23.820192 m)
Distance ≈ 96.948 m
So, the car traveled approximately 96.948 meters during this time.