1-D (I think) Speed to distance problem

[magnanimousto]

Homework Statement

A runner exerts a constant force to accelerate from rest to 2.0 m/s over a distance of 12 m.
Assuming the runner can keep up the same force, what total distance would be needed to ac-
celerate up from rest to 3.0 m/s?

Homework Equations

none given.

The Attempt at a Solution

The answer to the problem is 27. After quite some time spent thinking I tried solving first part for time using x= 1/2 vt and then plugging it into 2nd part. but that gives time as 12 and 2nd x as 18 so wrong. Dunno how to solve it

 

[SammyS]

Homework Statement

A runner exerts a constant force to accelerate from rest to 2.0 m/s over a distance of 12 m.
Assuming the runner can keep up the same force, what total distance would be needed to ac-
celerate up from rest to 3.0 m/s?

Homework Equations

none given.

The Attempt at a Solution

The answer to the problem is 27. After quite some time spent thinking I tried solving first part for time using x= 1/2 vt and then plugging it into 2nd part. but that gives time as 12 and 2nd x as 18 so wrong. Dunno how to solve it

Hello magnanimousto. Welcome to PF!

What kinematic equations do you know?

Alternatively, do you know the Work-Energy Theorem ?

 

[jackarms]

This is a constant acceleration problem, so you can use the basic kinematic equations for that case. Try solving for the acceleration first, and using that value for the next part.

If you don't know those equations, here's the one you should probably use -- they're very helpful to memorize though.
$v_{f}^{2}= v_{i}^{2} + 2aΔd$

 

[magnanimousto]

Thanks jackarms & SammyS.I do know kinematics equations, for some reason it never occurred to me that since the force is the same in both so must be the acceleration

 

[HalfLostAndSearching]

.

I would approach this problem by using the equations of motion, specifically the equation v^2 = u^2 + 2as. In this problem, u (initial velocity) is 0 m/s, v (final velocity) is 2.0 m/s, and a (acceleration) is unknown. We can rearrange the equation to solve for a: a = (v^2 - u^2)/2s. Plugging in the values, we get a = (2.0^2 - 0)/2(12) = 0.1667 m/s^2.

Now, we can use this acceleration value to solve for the total distance needed to accelerate to 3.0 m/s. Again, using the same equation, we have v^2 = u^2 + 2as, but this time v is 3.0 m/s and u is still 0 m/s. Plugging in a = 0.1667 m/s^2, we get s = (3.0^2 - 0)/2(0.1667) = 27 m. This is the total distance needed to accelerate from rest to 3.0 m/s.

In summary, the key to solving this problem is to use the equations of motion and to carefully consider the given information. By finding the acceleration in the first part of the problem, we can use it to solve for the total distance in the second part. It is important to carefully label and organize the given values in order to apply the equations correctly.