Solve for Velocity of Toy Car Accelerating with Rocket-Type Engine

[JasonRox]

Here is the question, it is very simple.

A toy car accelerates by means of a rocket-type engine for 20 m. If the acceleration during the burn is $$5 m/s^2$$ and the burn lasts for 3 s, determine the velocity of the car at the end of the burn.

It seems like the obvious one, and we use.

$$v=v_i + at$$

...and we get 15. This works for uniform acceleration, which it says it is.

They are using...

$$s = v_f t - 1/2 a t^2$$

They got that by differentiating from $$v_i$$, instead of the usual $$v_f$$

In the end, they get 14.2 m/s.

 

[ZapperZ]

Here is the question, it is very simple.

A toy car accelerates by means of a rocket-type engine for 20 m. If the acceleration during the burn is $$5 m/s^2$$ and the burn lasts for 3 s, determine the velocity of the car at the end of the burn.

It seems like the obvious one, and we use.

$$v=v_i + at$$

...and we get 15. This works for uniform acceleration, which it says it is.

They are using...

$$s = v_f t - 1/2 a t^2$$

They got that by differentiating from $$v_i$$, instead of the usual $$v_f$$

In the end, they get 14.2 m/s.

Er... you can't just switch around the "variable" in the problem. In

$$v=v_i + at$$

$$v_i$$ is a constant of integration that corresponds to the initial condition, where as v is THE variable, i.e. v=v(t).

This means that v = ds/dt. You cannot equate v_i as ds/dt because v_i IS A CONSTANT with respect to time. So that last equation is meaningless. This explains why you have a discrepancy in your answer.

Zz.

 

[JasonRox]

Hmm... you learn something everyday!

So, I guess I'm right about that. Also, this is high school physics, too. Mistakes are everywhere. Graphs are all messed up.

Anyways, I didn't realize that the variable cannot be a constant.