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MDCAT
Physics
2023

Physics · Forces and Motion

Work through this past-paper style MCQ, then read the full explanation. Practice more physics questions on mMCQ with adaptive practice and topic analytics.

Question

A train travels on a straight track passing signal A at 20 m/s. It accelerates uniformly at 2 m/s^2 and reaches signal B 100m further than A. At B the velocity of the train in m/s is

Options
  • A

    10

  • B

    20

  • C

    28

  • D

    56

Explanation

Using the equation of motion:


v² = u² + 2as


where:

v = final velocity (at signal B)

u = initial velocity (at signal A) = 20 m/s

a = acceleration = 2 m/s²

s = distance between signals A and B = 100 m


Substituting the values, we get:


v² = (20)² + 2(2)(100)

v² = 400 + 400

v² = 800

v = √800

v ≈ 28.28 m/s


Rounding to the nearest whole number, the velocity at signal B is approximately 28 m/s

Much too slow, incorrect calculation

Initial velocity, not final velocity

Using the equation of motion:

v² = u² + 2as

where:

v = final velocity (at signal B)

u = initial velocity (at signal A) = 20 m/s

a = acceleration = 2 m/s²

s = distance between signals A and B = 100 m

Substituting the values, we get:

v² = (20)² + 2(2)(100)

v² = 400 + 400

v² = 800

v = √800

v ≈ 28.28 m/s

Rounding to the nearest whole number, the velocity at signal B is approximately 28 m/s

Much too fast, incorrect calculation

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Tagged under Physics · Forces and Motion · 2023