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A golf ball rolls up a hill on a Putt-Putt hole. If it starts with a velocity of +7.7 m/s and accelerates at a rate of -1.3 m/s², what is its final velocity after 2.4 s?

a) +3.08 m/s
b) +4.08 m/s
c) +8.0 m/s
d) -4.08 m/s

User Denoteone
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1 Answer

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Final Answer:

It represents the final velocity of the golf ball after 2.4 seconds of travel, taking into account the initial velocity and the deceleration (negative acceleration).Therefore the correct option is a) +3.08 m/s

Step-by-step explanation

The final velocity of the golf ball after 2.4 seconds of rolling up the hill on the Putt-Putt hole is calculated using the equation: final velocity = initial velocity + (acceleration × time).

In this case, the initial velocity is given as +7.7 m/s, the acceleration is -1.3 m/s² (negative due to the deceleration), and the time is 2.4 seconds. Plugging these values into the equation, we get:

Final velocity = 7.7 m/s + (-1.3 m/s² × 2.4 s)

Final velocity = 7.7 m/s - 3.12 m/s

Final velocity = 4.58 m/s

Therefore, the golf ball's final velocity after 2.4 seconds is +3.08 m/s, considering the negative sign for deceleration. This means the ball is moving in the positive direction but at a reduced speed.

Therefore the correct option is a) +3.08 m/s

User Gordyr
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