The space shuttle, on reentry, slows its velocity from 565 km/hr to 496 km/hr in 15 seconds, then in the next 17 seconds it slows to 346 km/hr at touch down. What is its acceleration during this time frame?
-0.0019 km/s2
i know the answer is -0.019 km/s^2, but i don't know how i can get to the answer. can someone please help me
2007-01-17 05:55:29 · 3 answers · asked by star wars freak 2 in Science & Mathematics ➔ Physics
The acceleration is the rate of change of velocity: a=dv/dt. You need to find acceleration during the whole time (15+17=32sec.). The change in velocity during this time is 346-565=-219km/hr or -219/3600=-0.06083km/sec. So as stated above a=dv/dt=-0.06083/32=-0.0019km/s^2.
2007-01-17 06:59:24 · answer #1 · answered by Osve 1 · 0⤊ 0⤋
The change in velocity during the first period is -69 km/hr in 15 secs or -0.0192 km/sec over 15 seconds. (3600 seconds in hour)
The change in velocity for the second period is -150 km/hr or
-0.042 km/sec for 17 seconds.
The total change in velocity is -0.0612 over 32 seconds or
-0.00191 km/s2.
Note that you have two versions of the answer in your question.
2007-01-17 14:11:57 · answer #2 · answered by Mike1942f 7 · 0⤊ 0⤋
acceleration is change in velocity over time. You might get a different answer for the first 15 seconds than you get for the last 17. Calculate them separately.
2007-01-17 14:03:33 · answer #3 · answered by Anonymous · 0⤊ 0⤋