SOLUTION: A car travels on a straight, level road.
(a) Starting from rest, the car is going 48 ft/s (33 mi/h) at the end of 4.5 s. What is the car's average acceleration in ft/s2?
magnit
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Question 1046446: A car travels on a straight, level road.
(a) Starting from rest, the car is going 48 ft/s (33 mi/h) at the end of 4.5 s. What is the car's average acceleration in ft/s2?
magnitude: 10.66
in the direction of motion
(b) In 3.5 more seconds, the car is going 96 ft/s (65 mi/h). What is the car's average acceleration for this time period?
magnitude: 13.71
in the direction of motion
(c) The car then slows to 72 ft/s (49 mi/h) in 2.5 s. What is the average acceleration for this time period?
magnitude: 9.6
ft/s2
direction opposite to the direction of motion
(d) What is the overall average acceleration for the total time?
magnitude:
in the direction of motion
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A car travels on a straight, level road.
(a) Starting from rest, the car is going 48 ft/s (33 mi/h) at the end of 4.5 s. What is the car's average acceleration in ft/s2?
a = v/t = 48/4.5
a = 10.6667 ft/sec/sec
--------
(b) In 3.5 more seconds, the car is going 96 ft/s (65 mi/h). What is the car's average acceleration for this time period?
magnitude: 13.71
in the direction of motion
-----
a = (v2-v1)/t = (96 - 48)/3.5
a = 13.7143 ft/sec/sec
------
(c) The car then slows to 72 ft/s (49 mi/h) in 2.5 s. What is the average acceleration for this time period?
a = (v2-v1)/t = (96 - 72)/2.5
a = 13.7143 ft/sec/sec ********* ?????
magnitude: 9.6
ft/s2
direction opposite to the direction of motion
-------------
(d) What is the overall average acceleration for the total time?
magnitude:
in the direction of motion
---
Total time = 4.5 + 3.5 + 2.5 = 10.5 seconds
Terminal speed = 72 ft/sec
---
Avg a = 72/10.5 =~6.857 ft/sec/sec
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