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Question: If \({{x}}\) denotes displacement in time \({{t}}\) and \({{x = a cos t}}\), then acceleration is: ...

If x{{x}} denotes displacement in time t{{t}} and x=acost{{x = a cos t}}, then acceleration is:
(A)acost (B)−acost (C)asint (D)−asint  {{(A) a cos t}} \\\ {{(B) - a cos t}} \\\ {{(C) a sin t}} \\\ ({{D) - a sin t}} \\\

Explanation

Solution

Acceleration of a particle is the change in velocity of the particle with time. For finding acceleration, first of all find velocity (derivative of displacement) using formula, v⃗=dxdt{{\vec v = }}\dfrac{{{{dx}}}}{{{{dt}}}} and then find acceleration (derivative of velocity) using formula, a⃗=dv⃗dt{{\vec a = }}\dfrac{{{{d\vec v}}}}{{{{dt}}}}.

Complete step by step solution:
Given: x=acost{{x = a cos t}} where x = displacement and t = time
To find: acceleration
Acceleration of a particle is the time rate of change of its velocity. Acceleration is also defined as the change in velocity in unit time. It is a vector quantity denoted by a⃗{{\vec a}}.
Formula for acceleration, a⃗=dv⃗dt...(i){{\vec a = }}\dfrac{{{{d\vec v}}}}{{{{dt}}}}...{{(i)}}
Now, need to know the change in velocity
Velocity is the time rate of change of position of a particle in a particular direction. Velocity can also be defined as the time rate of change of displacement. It is a vector quantity denoted by v⃗{{\vec v}}.
Formula for velocity, v⃗=dxdt...(ii){{\vec v = }}\dfrac{{{{dx}}}}{{{{dt}}}}...{{(ii)}}
Now substituting the value of displacement, x=acost{{x = a cos t}} in formula (ii), we get
v⃗=ddt(acost) v⃗=−asint  {{\vec v = }}\dfrac{{{d}}}{{{{dt}}}}{{(a cos t)}} \\\ {{\vec v = - a sin t}} \\\
Now substituting the value of velocity in formula (i), we get
a⃗=ddt(−asint) a⃗=−acost  {{\vec a = }}\dfrac{{{d}}}{{{{dt}}}}{{( - a sin t}}) \\\ {{\vec a = - a cos t}} \\\
Thus, acceleration is −acost{{ - a cos t}}.

Therefore, option (B) is the correct choice.

Note: When a moving particle returns to its initial position then its displacement is zero and average velocity of the body is also zero but its average speed is not zero. In uniformly accelerated motion, average velocity over any interval of time is not equal to the instantaneous velocity. Acceleration can be either positive or negative. At any time t{{t}}, the direction of the motion is represented by velocity not by acceleration. For example, when an object is thrown vertically up then its velocity is directed upwards at any given time but acceleration is directed downwards.