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A rocket starts from rest and accelerates for 8 seconds at a rate of 3.6 m/s^2.

(A) What is its speed at the end of the 8 seconds?
(B) How far does the rocket travel during the FIRST FOUR SECONDS of the 8-second trip?
(C) How far does it travel during the last four seconds of the 8-second trip.

User Khasha
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1 Answer

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Final answer:

The speed of the rocket at the end of 8 seconds is 28.8 m/s. The rocket travels a distance of 28.8 meters during both the first four seconds and the last four seconds of the 8-second trip.

Step-by-step explanation:

(A) To find the speed of the rocket at the end of 8 seconds, we can use the equation:

v = vo + at

Where:

v is the final velocity,

vo is the initial velocity (0 m/s as the rocket starts from rest),

a is the acceleration (3.6 m/s²),

t is the time (8 seconds).

Substituting the values, we have:

v = 0 + 3.6 * 8 = 28.8 m/s

(B) To determine the distance traveled during the first four seconds, we can use the equation:

d = vo * t + 0.5 * a * t^2

Where:

d is the distance,

t is the time (4 seconds),

Substituting the values, we have:

d = 0 * 4 + 0.5 * 3.6 * 4^2 = 28.8 m

(C) To calculate the distance traveled during the last four seconds, we can use the same equation:

d = vo * t + 0.5 * a * t^2

Substituting the values:

d = 0 * 4 + 0.5 * 3.6 * 4^2 = 28.8 m

User Christopher Geary
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