Kinematics question - An automobile accelerating and braking

  • Thread starter rhoadsy74
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rhoadsy74
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kinematics question -- An automobile accelerating and braking

Homework Statement


According to recent test data, an automobile travels 0.250 mi in 19.9 s, starting from rest. The same car, when braking from 60.0 mi/h on dry pavement, stops in 146 ft. Assume constant acceleration in each part of the motion, but not necessarily the same acceleration when slowing down as when speeding up.
A. Find the acceleration of the car when it is speeding up in ft/s^2.
B. Find the acceleration of the car when it is braking in ft/s^2.
C. If its acceleration is constant, how fast (in mi/h) should this car be traveling after 0.250 mi of acceleration?
D. The actual measured speed is 70.0 mi/h; what does this tell you about the motion?
E. How long does it take this car to stop while braking from 60.0 mi/h?

Homework Equations


x=x(not)+V(not x)*t+(1/2)*a*t^2
v=V(not x)+a*t

The Attempt at a Solution



I am stuck on part A here is what i did to come to an answer that was wrong:

v=26.8224 meters/second V(not x)=0 and t=19.9 seconds

26.8224=0+a*19.9

I found that a= 4.43 feet/second^2.

I feel that the other questions in this problem is relying on part A.
 

Answers and Replies

  • #2
Doc Al
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I am stuck on part A here is what i did to come to an answer that was wrong:

v=26.8224 meters/second V(not x)=0 and t=19.9 seconds

26.8224=0+a*19.9

I found that a= 4.43 feet/second^2.

I feel that the other questions in this problem is relying on part A.
For part A you need to use your first equation, since you are given x and t.

Make sure your distances are in feet and the time in seconds, then your acceleration will be in the correct units of ft/s^2.
 

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