Question

A particle is moving along the curve \[x = a{t^2} + bt + c\], then the particle would be with uniform
A. rotation
B. velocity
C. acceleration
D. retardation

Answer 1

Hint: We must determine where the particle on the curve is uniform using the function of t that represents the equation of the curve in this question. When applying differentiation to the provided function, we eventually arrive at a constant function. Therefore, the constant function is uniform.

Formula Used: $\dfrac{{d({x^n})}}{{dt}} = n{x^{n - 1}}$

Complete step by step Solution: The equation of the curve is given by \[x = a{t^2} + bt + c\], the given equation is a function of t.
On differentiating the given function with respect to t we get
\[\dfrac{{dx}}{{dt}} = 2at + b\]
As a given function x will represent the distance. Then the rate of change in distance will represent the velocity.
On finding the first derivative, again it is a function of t. This is not a constant function and uniform also.
Again, the differentiation has to be done for the first derivative. So we get
\[\dfrac{{{d^2}x}}{{d{t^2}}} = 2a\]
The acceleration is represented by the second derivative, which follows the first derivative's representation of velocity. The second derivative is a uniform, constant function.
The particle is therefore travelling uniformly in all directions.

Hence, option C is correct one.

Note: The first derivative of distance will represent velocity and the first derivative of velocity will represent the acceleration or the second derivative of distance will represent the acceleration if the rate of change of distance is velocity and the rate of velocity is acceleration. Acceleration and velocity are both vector quantities.