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dimulka [17.4K]

3 years ago

14

Compare the results of applying the acceleration equation in the following 2 cases : (1) an object that goes from 0 to 10 m/s in

4 s , and (2) then goes from 10 m/s to 30 m/s in 8s.
a= (Vf-Vs)/t

1 answer:

erastovalidia [21]3 years ago

8 0

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Complete the statement about memory improvement.

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Read 2 more answers

A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.90 m/s2 . At 20.0s after blastoff, t

Brilliant_brown [7]

Answer:

A) 580m

B) 0 m/s

C) 9.8m/s^2

D) downward

E) 10.87s

F) 106.62 m/s

Explanation:

A) The distance traveled by the rocket is calculated by using the following expression:

$y=\frac{1}{2}at^2$

a: acceleration of the rocket = 2.90 m/s^2

t: time of the flight = 20.0 s

$y=\frac{1}{2}(2.90\frac{m}{s^2})(20.0s)^2=580m$

B) In the highest point the rocket has a velocity with magnitude zero v = 0m/s because there the rocket stops.

C) The engines of the rocket suddenly fails in the highest point. There, the acceleration of the rocket is due to the gravitational force, that is 9.8 m/s^2

D) The acceleration points downward

E) The time the rocket takes to return to the ground is given by:

$t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2(580m)}{9.8m/s^2}}=10.87s$

10.87 seconds

F) The velocity just before the rocket arrives to the ground is:

$v=\sqrt{2gy}=\sqrt{2(9.8m/s ^2)(580m)}=106.62\frac{m}{s}$

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a car going at 30 m/s undergoes an acceleration of 2 m/(s^2) for 4 seconds. how far did it travel while accelerating?

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