Question

The displacement x of a particle moving along a straight line at time t is given by $x={{a}_{0}}+{{a}_{1}}t+{{a}_{2}}{{t}^{2}}$. What is the acceleration of the particle?
A. ${{a}_{1}}$
B. ${{a}_{2}}$
C. 2${{a}_{2}}$
D. $3{{a}_{2}}$

Answer 1

Hint: Use definition of velocity as displacement per unit time and definition of acceleration as velocity per unit time. Use data in the given question. Apply concepts of calculus like derivation.

Complete Step-by-Step solution:
First understand what the question wants to say.
There is a particle whose displacement is x along a straight line at a time in a time t.
Value of x is given by
$x={{a}_{0}}+{{a}_{1}}t+{{a}_{2}}{{t}^{2}}$
Where \[{{a}_{0}},{{a}_{1}}and,{{a}_{2}}\]are constant.
The acceleration of the particle. First let us calculate velocity then acceleration.
Velocity: velocity is defined as displacement covered by the particle or object in a given time.
Which is given by,
$v=\dfrac{dx}{dt}$ -----------(1)
Acceleration: acceleration is defined as velocity of a particle in a given time.
Which is given by
$a=\dfrac{dv}{dt}$ _____________(2)
Put the value of x in equation (1). The value of x i.e. displacement is given in question
$v=\dfrac{d({{a}_{0}}+{{a}_{1}}t+{{a}_{2}}{{t}^{2}})}{dt}$
Use derivative w.r. to t and open parenthesis
\[v=\dfrac{d{{a}_{0}}}{dt}+\dfrac{d({{a}_{1}}t)}{dt}+\dfrac{d({{a}_{2}}{{t}^{2}})}{dt}\]
Since the derivative of constant is zero. Therefore the derivative of the first term is zero. Whereas derivative of constant cannot be zero if it is in a product with variable like t in the given question.
Therefore, constants will come out in second term as well as in the third term. And the variable goes under the derivative. So we get,
$v=0+{{a}_{1}}\dfrac{dt}{dt}+{{a}_{2}}\dfrac{d{{t}^{2}}}{dt}$
$v={{a}_{1}}+2{{a}_{2}}t$ -----------(3)
Now put value of velocity in equation (2)
$a=\dfrac{dv}{dt}=\dfrac{d({{a}_{1}}+{{a}_{2}}t)}{dt}$
Open parenthesis
\[a=\dfrac{d{{a}_{1}}}{dt}+{{a}_{2}}\dfrac{dt}{dt}\]
Again derivative of constant will be zero,
$a={{a}_{2}}$
Correct option is (B)

Note: Do not get confused between speed and velocity. Velocity is displacement per unit time and speed is distance per unit time. There is a difference between displacement and distance. Distance is measurement between two ends. And displacement is the path covered by an object. Limited derivatives are used in physics which you can memorize. One of them is used in this example.