An object starts from rest at point A and begins accelerating at a constant rate

Question

An object starts from rest at point A and begins accelerating at a constant rate of 1.5 m/s^2 until it reaches point C. At point C, the object begins to slow down and comes to a stop at point D, rests for 1.5 s and then accelerates at a constant rate back in the opposite direction until it reaches point B. Once the object starts moving, it takes the object 5.0 s to get from D back to B. The object stops accelerating at B, maintaining the same velocity from B to A.

The distance from A to B is 12.0 m. The distance from B to C is 15.0 m. The distance from C to D is 6.75 m.

A. Plot the trip on a position v. Time grap
B. plot the tip on a velocity v. Time graph

Explanation / Answer

A to C s= 0.5xaxt^2 12+15 = 0.5 x 1.5 xt^2 t= 6sec v=u+at = 1.5x6 = 9m/s u=0 v=9 m/s at t=6s C to D 6.75 = 9t -0.5axt^2 ; v=9-at at=9 6.75 = 9t -0.5x9t t=1.5 sec a=9/1.5 = 6m^2/s u=6m/s v=0m/s D to B 6.75+15=0.5 x a x t^2 t=5 s a= 1.74 m^2/s v=at =1.74x5= 8.7 m/s u=0 ;v=8.7 m/s from Bto A 12=8.7t t=1.38 sec A position vs t x = 0 27 33.75 33.75 12 0 t = 0 6 7.5 9 14 15.38 B) velocity vs time V= 0 9 0 0 8.7 8.7 t = 0 6 7.5 9 14 15.38

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