# Problem understanding simple constant acceeleration problem

• mkaa11
In summary, the problem asks a parachutist to calculate her final speed after falling 80 meters in 12 seconds. The problem states that her final speed is 25 m/s, but when Chet tried to check the answer, he found that it was different. He found that the answer is 24.7 m/s, which is the same as the answer that is given in the official answer provided by the physics department.
mkaa11
hi

i have an exam and i was trying solving old exams and i came upon this problem

1. Homework Statement

Due to air resistance, a 60- kg parachutist experiences an acceleration of 3.0 m/s2 while
falling in the air. If she falls 80 m in 12 seconds what is her final speed (in m/s)?

## Homework Equations

i know its a constant acceleration problem so
vf=vi+at
D=vit+0.5at^2

## The Attempt at a Solution

D is Y2-Y1 =(-80-0)=-80
by using 2nd equation vi=24.7
and then using 1st equation vf should be -11.3
so final speed should be 11.3
but when i checked the answer i found it different

here is the answer that i found
d=vt-1/2at^2
-80=12v-1/2(-3)(12)^2
v=24.7=25 m/s

Welcome to Physics Forums.

I'm not sure if your answer is 11.3, or 24.7, but I agree with the answer of 24.7 m/s

Since acceleration is constant, you can find the average speed by taking half of the sum of the initial and final speeds.
In 12 seconds at 3m/s/s, the initial speed would be 36 m/s less than the final speed, so you get $V_{avg}=\frac{1}{2}(V_f+(V_f-36))=V_f-18$
But the average speed is also the total distance divided by the total time so you get $V_{avg}=V_f-18=\frac{80}{12}$
and thus $V_f≈24.7$

Nathanael said:
Welcome to Physics Forums.

I'm not sure if your answer is 11.3, or 24.7, but I agree with the answer of 24.7 m/s

Since acceleration is constant, you can find the average speed by taking half of the sum of the initial and final speeds.
In 12 seconds at 3m/s/s, the initial speed would be 36 m/s less than the final speed, so you get $V_{avg}=\frac{1}{2}(V_f+(V_f-36))=V_f-18$
But the average speed is also the total distance divided by the total time so you get $V_{avg}=V_f-18=\frac{80}{12}$
and thus $V_f≈24.7$
Don't you think it's a little weird that the initial velocity was upward at 11.3 m/s? Do you think that whoever made up the problem didn't realize this?

Chet

Chestermiller said:
Don't you think it's a little weird that the initial velocity was upward at 11.3 m/s? Do you think that whoever made up the problem didn't realize this?

Oh wow, I didn't realize this either! Well, that doesn't make much sense...

Chestermiller said:
Don't you think it's a little weird
Was looking at it and thought I needed to take a nap. Tried pulling the rip cord at t=0, putting the chute into an updraft. Gave up.

Nathanael
Bystander said:
Was looking at it and thought I needed to take a nap. Tried pulling the rip cord at t=0, putting the chute into an updraft. Gave up.
I think that whoever made up the problem was taking a nap.

Chet

Nathanael
this was a short problem from last year exam (it's the official answer according to physics department)

so i am not doing something wrong?

I can't see how they got this answer. For one thing, the solution to their equation is -24.7, not + 24.7. The equation itself doesn't make sense to me.

Here's how I would have done it (with downward velocities, displacements, and accelerations treated as positive):
$$80 = v_i(12)+\frac{3}{2}(12)^2$$
v_i = -11.33 m/s

v_f = v_i + 3(12) = 24.7 m/s

Nathanael's method in post #2 is equally acceptable (maybe better), and gives the same answer.

Chet

Last edited:

## What is constant acceleration?

Constant acceleration is when an object's velocity changes at a constant rate over time. This means that the acceleration value remains the same throughout the entire motion of the object.

## How do I calculate constant acceleration?

To calculate constant acceleration, you need to use the formula a = (vf - vi)/t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time.

## What is the difference between constant acceleration and uniform acceleration?

Constant acceleration and uniform acceleration are often used interchangeably, but there is a slight difference. Constant acceleration refers to a motion where the acceleration remains the same for the entire duration of the motion, while uniform acceleration refers to a motion where the acceleration is constant and the velocity changes at a constant rate.

## How can I apply constant acceleration in real-life situations?

Constant acceleration can be observed in many real-life situations, such as a car accelerating on a straight road, a ball falling due to gravity, or a rocket launching into space. Understanding constant acceleration can help engineers and scientists design and improve various technologies and machines.

## How does air resistance affect constant acceleration?

In most cases, air resistance is negligible when it comes to constant acceleration problems. However, in situations where an object is moving at high speeds, air resistance can start to play a role and may need to be taken into consideration for more accurate calculations.

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