# Math Help - Intro Physics Problem. Need Help.

1. ## Intro Physics Problem. Need Help.

A model rocket rises with constant acceleration to a height of 3.2 m, at which point its speed is 26.0 m/s. (a) How much time does it take for the rocket to reach this height? (b) What was the magnitude of the rocket's acceleration? (c) Find the height and the speed of the rocket 0.10 after launch.

Need some help.

Thank You.

2. for this problem you need the kinematic equations. since it states that you have constant acceleration, you can use them. there are three you can use:
1) final velocity=initial velocity + acceleration(change in time)

2)initial velocity(change in time) + 1/2(acceleration)(change in time)^2= change in distance

3)(final velocity)^2=(initial velocity)^2+ 2(acceleration)(change in distance)

we will use the third one first.

consider 26 the final velocity and 0 the initial velocity(you'll have to assume this about the initial). also, 3.2 is the change in distance to that point. then you just solve for t.

then you use equation 1 with your new found acceleration, to find the change in time.

part c you probably want to find the speed first. use equation 1, use the same acceleration u used before, 0 is still the initial velocity and now the new change in time is .10. then just solve for the final velocity.

then for the height you can use either equation 2 or 3 to find the change in height by plugging in the time value, the acceleration, and the velocities.

hope this all helps, you will want to memorize these equations, they should be in your book.

3. First, we always consider what we know:

$x_0 = 0 m$

$x_f = 3.2 m$

$v_0 = 0 \frac ms$

$v = 26 \frac ms$

$a = ?$

$t = ?$

The first logical option would be to find t, we would first need the kinematic equation:

$x_f = x_0 + \frac 12 (v_0 + v)t$

$3.2 = \frac 12 (26)t$

$3.2 = 13t$

$t = 0.25 s$

Now, the next best option would be to find the acceleration, it is the only thing that we DON'T know, we can use the following equation:

$v = v_0 + at$

$26 = a(0.25)$

$a = 104 \frac{m}{s^2}$

Now, the last part is to find the speed of the rocket at time 0.10, so we take the following equation:

$v = v_0 + at$

And solve it for t=0.10 s:

$v = 0 + (104)(0.10)$

$v = 10.4 \frac ms$

For the height at time t=0.10, you can use this equation AGAIN:

$x_f = x_0 + \frac 12 (v_0 + v)t$

$x_f = \frac 12 (10.4)(0.10)$

$x_f = 1.04 m$

And that's it.

You were close if not right on most of them.

4. ## RE:

Thanks guys. I'm putting in a lot of hours trying to get the hang of this Physics stuff.

qbkr21

5. ## RE:

Aryth for the part of the problem in part (c) how did you figure to assume v_0 = 0? The problem states that the rocket rose with "constant" acceleration which made me believe from take-off it was something greater than zero.

EDIT: And why did you use 26 m/s as the velocity in part (c) instead of 10.56 m/s?

6. You assumed in your solution that the initial height was 0. If that is the case, then the rocket started on the ground. While it is on the ground it has no velocity. When it takes off with a constant acceleration, it builds a velocity over time. But it takes less than a fourth of a second to reach the 3.2 meters, which makes sense.

You're right about the velocity though, it should have been 10.4 or 10.56, whichever you got. That was an error on my part.