Question

5. (6 pts) Stan jumps out of an airplane with a parachute. After 34 seconds, he opens the parachute. The graph to the right shows Stan's vertical velocity versus time, in a coordinate system where down is positive. Velocity versus Time for a 70 kg Parachuter Stan opens parachute Jumps at t 0 Opens parachute at t 34s a 20 4--Stanjumps from airplane ! 0 10 20 30 40 50 60 time (s) The magnitude of Stan's initial acceleration is 9.81 m/s2, then changes with time due to air resistance. The magnitude of his acceleration just after 34 seconds is 9.40 m/s2 2 (a) Complete the chart. Include the correct signs and use 2 significant figures. time (s just after jump at 34 s 53 20 50 velocity m/s Sign of accleration 53 29-50-22 )Sketch a graph of Stan's acceleration versus time. Label the point where Stan jumps from the airplane and the point where Stan opens his parachute. Sketch of Acceleration versus Time 10 50 60 time, t (s) 10 20 4 10-

Answer 1

(a) In this, you values seems correct.

The sign of acceleration will be positive at t = 5 sec because speed is still increasing.

At t = 20 sec, a = 0 so, there is actually no sign. If you really want a sign, then it should be positive.

After that t = 34 sec, the acceleration is negative because speed is decreasing means Stan is deceleration

a = final velocity - initial velocity / time interval.

at t = 50 sec, the acceleration is again zero.

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The question tells that initial acceleration was 9.8 m/s² . so, the graph should start at 9.8.....rest everything is fine. Also, there is no acceleration till t = 34 sec, so extend your line till 34 seconds.