alleryStart learning →A train slows to a stop from a speed of 30.0 km/h in 8.00 s. What is its average acceleration in SI units?
A-3.75 m/s^2
B-1.04 m/s^2
C-2.08 m/s^2
D-0.104 m/s^2
Answer & Solution
Correct answer: B. -1.04 m/s^2
1. Identify the knowns: the initial velocity is +30.0 km/h and the final velocity is 0.
2. The change in velocity is 0 - 30.0 km/h = -30.0 km/h.
3. Convert to SI units: 30.0 km/h x (1000 m / 1 km) x (1 h / 3600 s) = 8.33 m/s.
4. Average acceleration is the change in velocity divided by the elapsed time.
5. Substitute the values: (-8.33 m/s) / (8.00 s) = -1.04 m/s^2.
6. The minus sign shows the acceleration opposes the motion, so this is a deceleration.
7. Dividing 30.0 by 8.00 without converting gives 3.75, which is the trap.
_Source: OpenStax College Physics (CC BY 4.0), Ch 2 "Kinematics", section 2.4 Acceleration_
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