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A train is moving at a speed of 35 m/s and decelerates to a stop in 12.6 s. What is the train's acceleration?

A. 2.78 m/s²
B. -2.78 m/s²
C. 0.79 m/s²
D. -0.79 m/s²

User Kairowa
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1 Answer

6 votes

Final answer:

The train decelerates from 35 m/s to 0 m/s in 12.6 s. By using the formula for acceleration (a = ∆v/∆t) the deceleration calculated is -2.78 m/s². Hence, the correct answer is B. -2.78 m/s². Option B is correct.

Step-by-step explanation:

To calculate the train's acceleration, we need to use the formula a = ∆v/∆t, where a is the acceleration, ∆v is the change in velocity, and ∆t is the change in time. The train decelerates from an initial velocity of 35 m/s to a final velocity of 0 m/s in a time span of 12.6 seconds. This means the change in velocity (∆v) is v_final - v_initial = 0 - 35 m/s = -35 m/s, indicating a decrease in speed.

To find the deceleration (which is negative acceleration), we use the formula: ∆v/∆t, which is: -35 m/s / 12.6 s = -2.78 m/s². Deceleration is negative because the train is slowing down, so the correct answer is B. -2.78 m/s².

To find the train's acceleration, we can use the formula:

acceleration = change in velocity / time taken

Given that the train's initial velocity is 35 m/s and it comes to a stop in 12.6 s, the change in velocity is 35 m/s and the time taken is 12.6 s. Thus, the acceleration is:

acceleration = 35 m/s / 12.6 s = - 2.78 m/s²

Therefore, the train's acceleration is - 2.78 m/s².

User Inkbug
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8.7k points