Distance/displacement, velocity, acceleration, equations, graphs MCQs

Q1. A passenger train of length 240 m is moving at 18 m/s. A man running at 6 m/s in the same direction takes how long to be completely overtaken by the train (i.e., the train fully passes him)?

• A.20 s
• B.10 s
• C.30 s
• D.40 s

Answer: A. 20 s

We are told: 240 m; 18 m; 6 m. To find: the unknown asked in the stem. Relation we use - man = 18 − 6 = 12 m/s. This is the equation that links the given quantities to the unknown (physics, chapter 'Motion'). Carrying out the arithmetic: man = 18 − 6 = 12 m/s → Time = 240 / 12 = 20 s. Why this is the right approach - Relative speed of train w.r.t. Putting it together the answer is A) 20 s.

Q2. A particle moving in a straight line with uniform acceleration has its velocity-square (v^2) vs position (x) graph as a straight line. The slope of this graph equals:

• A.2 a
• B.a
• C.a/2
• D.a^2

Answer: A. 2 a

Quick parse - a typical physics numerical. The data on the table: 2. We are after the quantity the stem asks for. Tool of choice - From v^2 = u^2 + 2 a x, the slope of a v^2 vs x graph is 2 a. Rearrange it for the unknown before substituting. Numbers in: From v^2 = u^2 + 2 a x, the slope of a v^2 vs x graph is 2 a. Common-sense check: The y-intercept gives u^2. Lock in option A) 2 a.

Q3. A car going at 90 km/h is brought to rest in 5 s by uniform braking. The deceleration and stopping distance are:

• A.5 m/s^2; 62.5 m
• B.5 m/s^2; 125 m
• C.18 m/s^2; 22.5 m
• D.25 m/s^2; 12.5 m

Answer: A. 5 m/s^2; 62.5 m