Kinematics refresher (straight-line motion): Which of the following are the standard constant-acceleration equations connecting displacement S, initial velocity u, final velocity v, acceleration a, and time t?
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Av = u + a t
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BS = u t + (1/2) a t^2
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Cv^2 − u^2 = 2 a S
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DAll of the above
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ENone of the above
Answer
Correct Answer: All of the above
Explanation
Introduction / Context:For motion along a straight line with constant acceleration, three core equations relate velocity, displacement, time, and acceleration. They are cornerstones in solving a wide range of problems from vehicle braking to projectile components and industrial automation timing.
Given Data / Assumptions:
- Acceleration a is constant in magnitude and direction.
- Variables: initial velocity u, final velocity v, displacement S, time t.
- Straight-line motion; vector signs handled consistently.
Concept / Approach:These equations are derived by integrating a = dv/dt, using v = ds/dt, and applying limits from initial to final states. Because a is constant, the integrals yield closed-form algebraic relations that avoid calculus during routine problem solving.
Step-by-Step Solution:
From a = dv/dt (constant), integrate: v = u + a t.Average velocity for constant a is (u + v)/2; therefore S = ((u + v)/2) * t.Eliminate v using v = u + a t to obtain S = u t + (1/2) a t^2.Eliminate t from v = u + a t and S = u t + (1/2) a t^2 to get v^2 − u^2 = 2 a S.Verification / Alternative check:Dimensional consistency: a t has velocity units; a t^2 has length units; the relation v^2 − u^2 = 2 a S has dimensions of velocity squared on both sides.
Why Other Options Are Wrong:Each of A, B, and C is correct individually; thus “All of the above” is the comprehensive choice. “None” is inapplicable.
Common Pitfalls:Using these equations when acceleration is not constant; mishandling signs when motion reverses direction during the time interval.
Final Answer:All of the above