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A car starts from xi = 11 m at ti = 0 and moves with the velocity graph shown be

ID: 1689624 • Letter: A

Question

A car starts from xi = 11 m at ti = 0 and moves with the velocity graph shown below. What is the object's position at: t= 2s, t= 3s, and t=4s? Does this car ever change direction? Answer yes or no. PLEASE SHOW ALL WORK!

Does this car ever change direction? Answer yes or no. PLEASE SHOW ALL WORK!
Does this car ever change direction? Answer yes or no. PLEASE SHOW ALL WORK!
A car starts from xi = 11 m at ti = 0 and moves with the velocity graph shown below. What is the object's position at: t= 2s, t= 3s, and t=4s? Does this car ever change direction? Answer yes or no. PLEASE SHOW ALL WORK!

Explanation / Answer

Distance is area under the velocity graph. So area at the end of 2s is 2*4 (area of rectangle) plus 1/2 2* (12-4) (area of triangle). So we have 8+ 8= 16. Add that to original position and you have 11+16= 27. At three you add the area of the triangle from 2 to 3 and you have 1/2 *4 *1= 2 so the position is 27+2= 29. At t= 3 the car starts changing direction (velocity becomes negative), so at t=4 we have 29- (1/2*4*1)= 27. Yes, the car changes direction at t=3.

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