Need help with this physics question please. Motion?

Question

A rocket is accelerating straight up from the earth's surface. at 1.15s after liftoff, the rocket clears the top of its launch platform, 63m above the ground. after an additional 4.75s, it is 1.00 km above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75s part of the flight and (b) the first 5.90s of the flight

thanks in advance, and please if possible can you explain it and write the whole solution in your answer? thanks a lot again!

Answer 1

Average velocity is displacement over time.  Displacement is a vector, but happily all of the motion in this problem is in one direction.  Call "up" positive.  Then:

For the first one:  t = 1.15 s, s = 63 m, so

v = s/t = 63m/1.15 s = 54.7826... m/s

As you've given it, the displacement has two significant digits (63) and the time has three (1.15 s), so the answer will have two significant digits:

v = 55 m/s

For the second part, t = 5.90 s and s = 1.00 km

v = s/t = 1.00 km/5.90 s = 0.169 km/s = 169 m/s

I'll let you figure out the significant digits.  This one isn't hard.