Acceleration Time graph of a parachutist who pulls his chute after jumping

It's wrong. C shows that there is constant acceleration then when the chute is pulled there is a large drop in acceleration but then rapidly rises.

So the answer should be D, since the acceleration increases then becomes constant ?

Actually no sorry I'm an idiot it is C sorry. Thought that on C it went back up to its original but looked second time and it rose up and stopped at 0. Others are incorrect since they show an increase in acceleration when its in the positive.

C is correct.

The jumper accelerates under constant gravitational acceleration (velocity increases) until his chute opens.

As the chute opens, he decelerates (curve goes negative) rapidly and attains a constant velocity at which time the acceleration is zero.

Sorry this is an old post, but what about if air resistance is taken into account?

Sorry this is an old post, but what about if air resistance is taken into account?

Sorry this is an old post, but what about if air resistance is taken into account?

i have a problem with this exact question because i thought the acceleration would decrease to zero, then stop and be constant. i dont fully understand why it continues further down? is it because of air resistance? If so please expound? but if the acceleration is negative, doesnt that mean its moving in the opposite direction, and does that even make sense then?

i have a problem with this exact question because i thought the acceleration would decrease to zero, then stop and be constant. i dont fully understand why it continues further down? is it because of air resistance? If so please expound? but if the acceleration is negative, doesnt that mean its moving in the opposite direction, and does that even make sense then?

Initially, the acceleration and velocity are in the same direction i.e. pointing down toward the Earth. When the parachute is opened at t = 2.5s, air resistance increases rapidly. The air resistance is much greater than the weight of the parachutist, so the direction of acceleration changes rapidly from downwards to upward (i.e. against the motion of the parachutist). As a result, the motion of the parachutist slows down. When the speed of the parachutist slows down, air resistance decreases. The magnitude of the acceleration will also decrease and reach zero eventually when the terminal velocity is reached.

Note that a number of the helpers mention that negative acceleration means deceleration. This is NOT always the case. The interpretation of negative acceleration should take into account the direction of the velocity.

This is clearly if there is no air resistance. If there was to be air resistance, wouldnt the resultant force begin to decrease, causing a decreasing value of acceleration until it becomes zero at terminal velocity. Then the graph will continue after 2.5 like at C?
Please correct me if i am wrong

... If there was to be air resistance, wouldnt the resultant force begin to decrease, causing a decreasing value of acceleration until it becomes zero at terminal velocity. Then the graph will continue after 2.5 like at C? ...