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A snowmobile is traveling at 150 mi/hr. Over the course of 18.0 seconds, the driver decelerates to rest. What is the acceleration of the snowmobile during this time?

A) 16.67 mi/hr^2
B) 20.83 mi/hr^2
C) 8.33 mi/hr^2
D) 10.42 mi/hr^2

1 Answer

4 votes

Final answer:

The snowmobile decelerates from 150 mi/hr to 0 mi/hr over 18 seconds, resulting in a deceleration of -30 mi/hr², which is not one of the provided options (A, B, C, D). Thus, there seems to be a mistake as the correct deceleration rate is not listed.

Step-by-step explanation:

To calculate the acceleration of the snowmobile, we can use the formula a = Δv/Δt, where Δv is the change in velocity and Δt is the change in time. Since the snowmobile is decelerating, we expect a negative acceleration value. The snowmobile goes from 150 mi/hr to rest, which is 0 mi/hr, over the course of 18.0 seconds.

First, we convert the initial velocity to the same units used for acceleration (mi/hr2). Since there are 3600 seconds in an hour, we can express 150 mi/hr as 150 mi/hr * (1 hr / 3600 s) to get the speed in mi/s and then multiply by the time in seconds to get the change in velocity.

Δv = 150 mi/hr = 150 / 3600 mi/s

Now we calculate the acceleration:

a = Δv/Δt = (0 - (150/3600)) mi/s / 18 s = -0.008333... mi/s2

The minus sign indicates deceleration. To get the units back to mi/hr2, we multiply by 36002 since we need to square the conversion factor to account for the square in the units.

a = -0.008333... mi/s2 * 36002 = -30 mi/hr2

None of the options (A, B, C, D) are correct. The actual deceleration is -30 mi/hr2, which means the snowmobile's velocity is decreasing by 30 mi/hr each second.

User Hamed Minaee
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