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Question: The displacement (in meter) of a particle moving along the x-axis is given as \(x = 18t + 5{t^2}\)...

The displacement (in meter) of a particle moving along the x-axis is given as
x=18t+5t2x = 18t + 5{t^2}.
Calculate
A.) The instantaneous velocity at t = 2 sec
B.) Average velocity between t = 2 sec and t = 3 sec.

Explanation

Solution

Hint: Instantaneous velocity is calculated by taking the derivative of displacement with respect to time at a particular value of time whereas the average velocity between two points is calculated by dividing velocity with time.

Complete step by step solution:
Instantaneous velocity of a moving body is defined as the velocity of the body at a particular instant of time during its motion. Velocity of the body may keep changing while it is in motion. Instantaneous velocity tells us the speed of the object at a particular instant of time.
It can be calculated by taking the time derivative of displacement at a particular instant of time.

Average velocity of a body between two points is simply the total displacement of that body between two points divided by the total time taken to get from one point to another.

In this question, we are given the expression for displacement which is x=18t+5t2x = 18t + 5{t^2}

In order to calculate instantaneous velocity at t = 2sec, we first take the time derivative of displacement as follows.

v=dxdt=ddt(18t+5t2)=18+10t  {\text{v}} = \dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left( {18t + 5{t^2}} \right) = 18 + 10t \\\

This is the general expression of instantaneous velocity at any time t. We need its values for t = 2 sec which is

v=18+10(2)=18+20=38ms1{\text{v}} = 18 + 10\left( 2 \right) = 18 + 20 = 38m{s^{ - 1}}

Which is the required answer.

The average velocity between t = 2 sec and t = 3 sec can be calculated as follows.

Displacement at t = 2 sec, x=18(2)+5(2)2=36+5(4)=36+20=56mx = 18\left( 2 \right) + 5{\left( 2 \right)^2} = 36 + 5\left( 4 \right) = 36 + 20 = 56m

Displacement at t = 3 sec,

x=18(3)+5(3)2=54+5(9)=54+45=99mx = 18\left( 3 \right) + 5{\left( 3 \right)^2} = 54 + 5\left( 9 \right) = 54 + 45 = 99m

Average velocity,

vav=x2x1t2t1=995632=43ms1{{\text{v}}_{{\text{av}}}} = \dfrac{{{x_2} - {x_1}}}{{{t_2} - {t_1}}} = \dfrac{{99 - 56}}{{3 - 2}} = 43m{s^{ - 1}}

Which is the required answer.

Note: There is no derivative calculated in case of average velocity. It is simply the displacement divided by time while instantaneous velocity is basically the average velocity with a limit of time difference approaching zero.