# Thread: Calculate acceleration with time and distance?

1. ## Calculate acceleration with time and distance?

In the 74s after lift-off, the shuttle Challenger travels 36km. Assuming constant acceleration, calculate the acceleration of the shuttle in m/s^2.

G:
t = 74s
d = 36km

Which equation would I use to solve this? None of the equations I have will work, and I'm not sure which kind of velocity to calculate (instantaneous, average, initial, final). Everything I tried so far failed. The correct answer (according to my sheet) is 13m/s^2.

In the 74s after lift-off, the shuttle Challenger travels 36km. Assuming constant acceleration, calculate the acceleration of the shuttle in m/s^2.

G:
t = 74s
d = 36km

Which equation would I use to solve this? None of the equations I have will work, and I'm not sure which kind of velocity to calculate (instantaneous, average, initial, final). Everything I tried so far failed. The correct answer (according to my sheet) is 13m/s^2.
assuming the shuttle started from rest at t = 0 ...

$\displaystyle d = \frac{1}{2}at^2$

solve for $\displaystyle a$

remember that $\displaystyle d$ must be in meters

3. Thanks for the help, got it right finally.

Though, the second part of the question asks to find the speed of the shuttle in km/h. I did the following.

S = a(t)
S = 13m/s^2(74s)
S = 962m/s
S = 962(.001km)(.0003h)
S = 962(3.33km/h)
S = 3203.46km/h.

The answer on the page however, says 3500km/h. Don't know if this was due to rounding or not, though I doubt that based on the high inaccuracy.

I also tried this.

S = d / t
S = 36km / 74s
S = 0.49km/s
S = 0.49km/(0.0003h)
S = 0.000147km/h

I thought for sure this one would work, but I guess not.

What would be a good equation for calculating the shuttle's speed?

4. $\displaystyle a \approx 13.15 \, \, m/s^2$

$\displaystyle v = at \approx 973 \, \, m/s \, \, \approx 3500 \, \, km/hr$