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Difficulty: MediumKinematics and Linear Motion

An electric train traveling along a straight track uniformly slows down from a speed of 30 m/s30\text{ m/s} to 10 m/s10\text{ m/s} over a distance of 200 m200\text{ m}. What is the magnitude of the deceleration of the train in m/s2\text{m/s}^2?

Answer: 2 m/s^2

Answer

The magnitude of the deceleration is 2 m/s22\text{ m/s}^2.
Using the kinematic equation v2=u2+2asv^2 = u^2 + 2as, substituting v=10 m/sv = 10\text{ m/s}, u=30 m/su = 30\text{ m/s}, and s=200 ms = 200\text{ m} gives 100=900+400a100 = 900 + 400a, which simplifies to 400a=−800400a = -800, yielding a=−2 m/s2a = -2\text{ m/s}^2. The magnitude of deceleration is 2 m/s22\text{ m/s}^2.

Step-by-Step Solution

1
Identify the given kinematic variables.
Initial velocity u=30 m/su = 30\text{ m/s}, final velocity v=10 m/sv = 10\text{ m/s}, displacement s=200 ms = 200\text{ m}.
Choosing the appropriate equation of motion requires knowing which variables are given and which is unknown.
2
Apply the third equation of motion relating initial velocity, final velocity, acceleration, and distance.
v2=u2+2asv^2 = u^2 + 2as
This formula connects uu, vv, aa, and ss without needing time tt.
3
Substitute the given values into the equation and solve for acceleration aa.
(10)2=(30)2+2(a)(200)  ⟹  100=900+400a  ⟹  400a=−800  ⟹  a=−2 m/s2(10)^2 = (30)^2 + 2(a)(200) \implies 100 = 900 + 400a \implies 400a = -800 \implies a = -2\text{ m/s}^2.
Performing algebraic operations to isolate the acceleration parameter.
4
State the magnitude of the deceleration.
The magnitude of deceleration is 2 m/s22\text{ m/s}^2.
Deceleration represents the rate of speed reduction, which corresponds to the magnitude of negative acceleration.

Key Concept

Uniformly Accelerated Motion Equations
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Topics

Kinematics and Linear Motion

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