Question

A particle moves in a straight line. It's position (in mts) as function of time is given by
\[\text{x = a}{{\text{t}}^{2}}+\text{b}\]
What is the average velocity in time interval \[t=3s\] to \[t=5s\] in \[m{{s}^{-1}}\]. (where a and b are constants and \[a=1m{{s}^{-2}}\]and \[b=1m\]).

Answer 1

Hint: In general as we know average velocity can be found by dividing total distance from time. So we can find the total distance travelled by particle from 3s to 5s. We just need to find the position of the particle at 3sec and 5sec by putting the value of t as 3 and 5 in the equation. Then we just need to subtract it to get the distance.

Complete step-by-step answer:
Velocity can be explained as the rate of change of displacement. It is a vector quantity i.e.,
\[v=\dfrac{dx}{dt}\]…………………………….……. 1
Whereas average velocity can be defined as the ratio of the total distance travelled by the particle to the time taken for that total distance travelled which can be mathematically expressed as
\[{{\text{v}}_{\text{avg}}}\text{=}\dfrac{\text{Total distance travelled}}{\text{time taken}}\]
i.e., \[{{v}_{avg}}=\dfrac{\vartriangle x}{\vartriangle t}\]
Given data
\[x=a{{t}^{2}}+b\]
t is from 3 sec to 5 sec
Where as a and b are constants
\[a=1m{{s}^{-2}}\] and \[b=1m\]

Since the distance travelled by the particle as a function of t \[\vartriangle x\] i.e., total distance travelled in the time limits can be calculated by putting & values in the function given and subtracting the final distance and the initial distance.
i.e., \[x(t)=a{{t}^{2}}+b\]
\[x(3)=(1){{(3)}^{2}}+1=10\]
\[x(5)=(1){{(5)}^{2}}+1=26\]
Therefore \[\vartriangle x=x(5)-x(3)=26-10=16\]
Therefore, the total distance travelled by the particle is 16 m.
From the given time interval we can find the total time taken as \[t=5-3=2\sec \]
Therefore, total time taken is sec from the above definition of average velocity.
Average velocity \[=\dfrac{\vartriangle x}{\vartriangle t}=\dfrac{16}{2}=8m{{s}^{-1}}\]
Therefore, the required average velocity is \[8m{{s}^{-1}}\]

Note: We should take care while calculating average velocity which is different from that of velocity (or) velocity at a particular point of time. This problem can also be solved by finding out the velocities at time \[t=3\]sec and \[t=5\]sec and taking the average & these two velocities. Using the given displacement (function of time).
Refer equation 1 for better understanding.