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?
1. -0.0019 km/s2
2. -0.2 km/s2
3. -0.06 km/s2
4. -219 km/s2
2006-10-25 06:29:51 · 3 answers · asked by maconheira 4 in Science & Mathematics ➔ Mathematics
From the answer choices, apparently "this time frame" refers to both instances of deceleration
V'= 565 km/hr; V"= 346 km/hr: V"-V'= -219km/h, or -219,000m/h
Deceleration occurred for (15 + 17 =) 32 seconds.
Since a= (V"-V')/t, a= -219,000m/h divided by 32s = roughly -6843.75 m/h/s (an ungainly unit, and not one of the answer choices).
Remembering there are 3600 seconds per hour, we multiply -6843.75 m/h/s * 1h/3600s to get -1.90104 m/s/s (or -1.90104 m/s^2) -- which, although neat, isn't among the answer choices.
Remembering there are 1000m/km, we multiply by 1km/1000m to get roughly -0.0019 km/s^2 -- which, conveniently, is among the possible answers.
Average acceleration is -0.0019 km/s^2.
2006-10-25 07:00:13 · answer #1 · answered by wireflight 4 · 0⤊ 0⤋
it is 219km/3600s*32s=-0.0019km/s^2
2006-10-25 13:39:17 · answer #2 · answered by raj 7 · 0⤊ 0⤋
very VERY! fast
2006-10-25 13:32:05 · answer #3 · answered by Anonymous · 0⤊ 1⤋