A car starts from rest, then accelerates at a constant rate over a distance of 72 m. It then immediately decelerates at a constant rate over a distance of 141 m. The entire trip lasts a total duration of 13.7 s. What were the magnitudes of the car s accelerations for the speedup and slowdown stages respectively? O 6.71 m/s^2, then 3.43 m/s^2 5.26 m/s^2, then 10.29 m/s^2
Displacement, Velocity and Acceleration
In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.
Linear Displacement
The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.
A car starts from rest, then accelerates at a constant rate over a distance of 72 m. It then immediately decelerates at a constant rate over a distance of 141 m. The entire trip lasts a total duration of 13.7 s. What were the magnitudes of the car s accelerations for the speedup and slowdown stages respectively?
![A car starts from rest, then accelerates at a constant rate over a distance of 72 m. It then immediately decelerates at a constant rate
over a distance of 141 m. The entire trip lasts a total duration of 13.7 s. What were the magnitudes of the car s accelerations for the
speedup and slowdown stages respectively?
O 6.71 m/s^2, then 3.43 m/s^2
5.26 m/s^2, then 10.29 m/s^2
2.27 m/s^2, then 2.27 m/s^2
10.29 m/s^2, then 5.26 m/s^2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6592bf53-b036-46fe-b1f5-a73697a8faa9%2Fb3cbd6a4-50d2-4bfc-809c-9cf8db0f4b47%2Fjboaxx8_processed.jpeg&w=3840&q=75)
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Given
Initial velocity u = 0
Distance travelled during acceleration = 72 m
Distance covered during deacceleraration = 141 m
Total time taken = 13.7 s
Area under velocity time graph represents displacement.
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