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anonymous
 one year ago
What is the average acceleration of a jet ski that goes from rest to 10 m/s west in 4 s? Take the direction west to be positive, and east to be negative.
A.
2.5 m/s2
B.
2.5 m/s2
C.
0.4 m/s2
D.
0.4 m/s2
anonymous
 one year ago
What is the average acceleration of a jet ski that goes from rest to 10 m/s west in 4 s? Take the direction west to be positive, and east to be negative. A. 2.5 m/s2 B. 2.5 m/s2 C. 0.4 m/s2 D. 0.4 m/s2

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taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.2Acceleration is just the change in velocity divided by the change in time so you just have to take the 10m/s divided by 4s Since west is positive and the jet ski is going west your answer will be positive.

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.2Yes it would be A

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sweet if you help me with one more i'll give you a metal

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Which of the following has no acceleration? A. a satellite orbiting at a constant speed B. a person running once around a track at a constant speed C. a car making a turn at a constant speed D. a person running 100 m straight across a field at a constant speed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@taramgrant0543664 ?

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.2It would be D because acceleration exists for the rest because there was a change in direction but not in this case it was a straight run

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Think of acceleration as \[\frac{ \Delta v }{ \Delta t }\] so the change in velocity over time

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yes, D is right, as your speed is at constant a = 0

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1As tara mentioned, there is no change in direction and if you think of velocity, which is \[\frac{ \Delta d }{ \Delta t }\]
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