Calculus Position Velocity Acceleration Test 1

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[sachinrath]

at t = 9, a is zero, that does not mean breaks are applied as when breaks are applied acceleration decreases, does not become zero.

[johnnolen]

No doubt, one should pause the video after reading this question. While I agree that your procedure is correct for determining the answers to parts a, b, and c; there is no way on God's Earth that s(t) represents the position for a car! I solved the problem and then asked myself the question, "What's going on such that the velocity and acceleration are each initially negative?" Is the car backing up with increasing speed? That's not reasonable. The remaining alternative is that s'(t) represents a closing speed since "the car" experiences its minimum velocity at t = 9 seconds but has positive acceleration thereafter until it stops at t = 15 seconds. (Velocity and acceleration having opposite signs indicates the vehicle is slowing down. Therefore, s'(t) must represent a rate of closure, which makes sense if approaching a stop sign.) Even taking this into account, the minimum velocity [maximum closure rate], v(t = 9 seconds) = - 3888 m/s, nearly twice as fast as the fastest vehicle speed ever recorded; the 3 October 1967 flight of the X-15, top speed 4520 miles per hour.