Question

A particle moves along a parabolic path $y = 2x - {x^2} + 2$in such a way that the x-component of the vector remains constant ($5\,m/s$). Find the magnitude of acceleration of the particle.

Answer 1

The difference between the final and initial velocity with respect to time is called acceleration. And it is denoted by ‘a’ and the SI unit is $m{s^{ - 2}}$.Acceleration is a vector quantity, it means it has direction as well as magnitude. There are two types of acceleration: uniform acceleration and non-uniform accelerations.

Complete step by step answer:
Given equation
$y = 2x - {x^2} + 2$............…(1)
Velocity along x-direction,
${v_x} = 5\,m/s$............…(2)
Component of the velocity vector along a particular direction is equal to the time rate of change of displacement along that direction.
So the velocity of component of x and y are express by
${v_x} = \dfrac{{dx}}{{dt}}$, ${v_y} = \dfrac{{dy}}{{dt}}$
Now differentiate the equation (1) with respect to time
$y = 2x - {x^2} + 2$
$\Rightarrow \dfrac{{dy}}{{dt}} = 2\dfrac{{dx}}{{dt}} - 2x\dfrac{{dx}}{{dt}} + 0$
Above equation can be written as
${v_y} = 2{v_x} - 2x{v_x}$..........…(3)
Put the value of ${v_x}$ from equation (2)
${v_y} = 2(5) - 2x(5)$
$\Rightarrow {v_y} = 10 - 10x$.........…(4)

As the component of velocity along the x-axis is not changing with time, so resultant acceleration is zero. So we differentiate equation number 3 with respect to time.
${a_x} = 0$
$\Rightarrow {a_y} = \dfrac{d}{{dt}}({v_y})$
$\Rightarrow {a_y} = \dfrac{d}{{dt}}(10 - 10x)$
$\Rightarrow {a_y} = 0 - 10\dfrac{{dx}}{{dt}}$
$\Rightarrow {a_y} = - 10{v_x}$
Using the value of ${v_x}$ from equation 2
${a_y} = - 10(5)$
$\therefore {a_y} = - 50\,m/{s^2}$

Hence, acceleration of the particle is $50\,m/{s^2}$, directed along negative y-direction.

Note: Force is the product of mass and acceleration. Application acceleration improves the application performances using techniques like compression, caching and transmission control protocol. When the velocity of any moving is changed it is said to be change in acceleration. Change in acceleration may be positive or negative.