# Acceleration, displacement, Kinematics Question

• ballerina_tee
In summary, Superwoman is hovering above the ground when a person free-falling goes by her at a terminal velocity of 140 km/h. Unfortunately, the parachute does not open, but luckily Superwoman is around. If it takes her 1.9 seconds to realize the person is in distress, she would need to accelerate at a rate of 256 m/s^2 in order to catch the parachutist just before he hits the ground 1000 m below.
ballerina_tee
[SOLVED] Acceleration, displacement, Kinematics Question

1. Superwoman is hovering above the ground when a person free-falling goes by her at a terminal velocity of 140 km/h. Unforunately, the parachute does not open. Forunately, Superwoman is around. If it takes her 1.9 s to realize the person is in distress, what mus her acceleration be if she is to catch the parachutist just before she hits the ground 1000 m below?

2. I think you use the equation d= v*t + 0.5 a*t^2 and rearrange for a which i got as a=d-v*t/ t^2

3. So I plugged in the numbers given in the question

a=d-v*t/ t^2
= (1000m) -(38.89) (1.9) / 1.9^2
=926.1/3.61
=256 m/s^2

That doesn't look right to me. The answer is supposed to be 3.53 m/s^2. I have no clue what I'm doing wrong! =( Please help! TIA.

I'm not sure why you're trying to find the acceleration of the parachutist. It makes no sense. The parachutist isn't accelerating, because he has reached terminal velocity. I don't understand why you have the time as 1.9 s.

What you need to do is figure out how much time it will take the parachutist to reach the ground starting from when he passes Superwoman. Once you know this, you can figure out what acceleration Superwoman needs to reach the parachutist before they get to the ground.

Also, you seemed to have dropped the 0.5 when you were rearranging your equation.

I did what you suggested and according to my calculations, I got 51.42 seconds.

I used this equation and solved for t.
d=1/2(v2+v1)*t

I just don't know what equations to use.

I got rid of the 0.5 by multiplying by 2. I forgot to add that.

a=d-v*t/ t^2 *(2)

I did what you suggested and according to my calculations, I got 51.42 seconds.

This double what I think you should be getting. What you're using in your equation is the average velocity (which technically should have worked for this so I suspect you just made a math error). But anyway, the velocity isn't changing since it's constant, so you can just use v = d/t which gives t = d/v.

You can use the equation you had in your first post to find the acceleration of Superwoman. You know d, since Superwoman catches the parachutist just before he hits the ground. Since she was "at rest" when the parachutist went by, her initial velocity is zero, so your equation becomes simpler. What you have to figure out is t. Once you have that, you just solve for a.

So, if it takes the parachutist a certain time to reach the ground from 1000 m, and Superwoman waits 1.9 seconds before she starts moving, how much time is left before the parachutist hits the ground? This is what you need to find.

Ah, I feel so dumb.

So, to find t, you use t=d/v (1000/140)? I understand what the question is asking of me but I'm totally lost in what equations I should be using to solve the problem. The question isn't difficult but every time I try to solve, I get a number that is way off.
I'll try it again.

Okay, for the 10000th time lol,

I did t=d/v (1000m/38.89m/s) therefore t = 25.7 s

I plug that into the equation to find acceleration
a=d-v*t/ t^2 *(2)
= (1000m) - (0m/s)(25.7)/ 25.7^2 *2
= 1000/ 660.49 *2
= 1.5 X 2
= 3.02 m/s^2

Would this be correct?

The back of the textbook says the answer should be 3.53 m/s^2. Close Enough I guess.

Your time isn't quite correct. Remember that Superwoman waits 1.9 seconds before she starts moving. So if it takes the parachutist 25.7 seconds to reach the ground, and Superwoman waits 1.9 seconds, how much time is left? (i.e. how much time does Superwoman now have to get to the parachutist?) This is the time you need to find acceleration, it's NOT just the 25.7 s.

The back of the textbook says the answer should be 3.53 m/s^2. Close Enough I guess.

Nope. Your answer is not close enough, it's wrong. Fix the time and you will have it truly right.

Last edited:
Simple...25.7-1.9=23.8 seconds.

I used 23.8 instead of 25.7 into the equation and got 3.5 seconds

yay! Thank you!

You're welcome.

## 1. What is acceleration?

Acceleration is the rate of change of velocity over time. It is a vector quantity, meaning it has both magnitude and direction. It is typically measured in meters per second squared (m/s²).

## 2. How is acceleration calculated?

Acceleration can be calculated by dividing the change in velocity by the change in time. The formula for acceleration is a = (vf - vi) / t, where vf is the final velocity, vi is the initial velocity, and t is the time interval.

## 3. What is displacement?

Displacement is the change in position of an object. It is a vector quantity, meaning it has both magnitude and direction. It is typically measured in meters (m).

## 4. How is displacement related to acceleration?

Displacement is related to acceleration through the equation d = vi*t + 1/2*a*t², where d is displacement, vi is initial velocity, t is time, and a is acceleration. This equation is derived from the definition of acceleration as the rate of change of velocity over time.

## 5. What is kinematics?

Kinematics is the branch of physics that studies the motion of objects without considering the forces that cause the motion. It involves the study of displacement, velocity, acceleration, and time to describe the motion of objects.

• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
761
• Introductory Physics Homework Help
Replies
4
Views
812
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
38
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
993