Finding vertical displacement

Question

I am being asked to find the distance a shuttle travels upward after a given amount of time. I know that time passed is 79s, the rate of acceleration is 6.244 m/s^2, and the speed at 79s is 493.276 m/s. However I can't figure out what equation to use to find the altitude at that time, 79 s. Please help! Thank you.

Answer 1

Motion under constant acceleration is given by $\ddot{x}(t) = a.$ Integrating twice gives

$$x(t) = x_0 + v_0t + \frac{at^2}{2}.$$

If you set your initial height to $x_0 = 0$ and assume that $v_0 = 0$ as well then $x(79)$ follows easily.

Answer 2

I presume the space shuttle is starting from rest at the launch pad at time $t=0$, correct?

I also assume the acceleration is constant.

Speed after 79 s

Velocity after a given time $t$ is related to the initial velocity and the acceleration through

$v = v_0 + a t$

where $v$ is the velocity at time $t$, $v_i$ is the initial velocity, $a$ the acceleration. If you plug in the units, you will see all terms have units $[m/s]$. That is always the first check. For more information see Wikipedia .

As $v_0=0$ we have $v = a t = 3.244*79 = 493.3 \ m/s$ Note that we only have 4 significant digits in the acceleration so can only find the velocity to 4 significant digits.

Displacement

Now for the displacement, use the formula

$x = v_0 t + \frac{1}{2} a t^2$

also from Wikipedia. In your case it simplifies to $x =\frac{1}{2} a t^2$. As you can see, you don't need to know the velocity after 79s.