Question

The position of an object moving along X-axis is given by $x=a+b{ t }^{ 2 }$ where a= 8.5 m, b= $2.5m{ s }^{ -2 }$ and t is measured in seconds. What is its velocity at t= 0 s and t= 2.0 s. What is the average velocity between t = 2.0 s and t= 4.0 s?

Answer 1

To solve the first problem, first find the position of the object at t= 0 s by substituting the values in the equation for the position given. Then, similarly find the position at t= 2 s and t= 4 s. For the first part of the question, find the displacement of the particle from t= 0 s and t=2 s and find the difference between these two times. Substitute these values in the formula for velocity which gives relation between displacement, time and velocity. This will give the velocity between t= 0 s and t= 2 s. For the second part of the question, follow the same steps as the first part but for t= 2 s and t= 4 s. This will give the average velocity between t= 2 s and t= 4 s.

Complete answer:
Given: a= 8.5 m
b=$2.5m{ s }^{ -2 }$
Position is given as,
$x=a+b{ t }^{ 2 }$ …(1)
At t= 0 s, equation. (1) becomes,
${x}_{2}=a+b{( 20)}^{ 2 }$
$\rightarrow{ x}_{2} =a+0$
Substituting the values in above equation we get,
${x}_{2}= 8.5 + 0$
$\Rightarrow {x}_{2}= 8.5 m$
At t= 2 s, equation. (1) becomes,
${x}_{2}=a+b{( 2 )}^{ 2 }$
$\rightarrow{ x}_{2} =a+4b$
Substituting the values in above equation we get,
${x}_{2}= 8.5 + 4 \times 2.5$
$\Rightarrow {x}_{2}= 18.5 m$
Similarly, at t= 4 s, equation. (1) becomes,
${x}_{3}=a+b{( 4) }^{ 2 }$
$\rightarrow {x}_{3} =a+16b$
Substituting the values in above equation we get,
${x}_{3}= 8.5 + 16 \times 2.5$
$\Rightarrow {x}_{3}= 48.5 m$
Displacement from t= 0 s to t= 2 s is given by,
${S}_{1}= {x}_{2}- {x}_{1}$
Substituting the values in above equation we get,
${S}_{1}= 18.5 – 8.5$
$\Rightarrow {S}_{1}= 10 m$
Time taken by the object to move from ${x}_{1}$ to ${x}_{2}$ is given by,
${ t }^{ ' }=2-0$
$\Rightarrow { t }^{ ' }=2 s$
Formula for velocity is given by,
$Velocity= \dfrac {Displacement}{time}$
Therefore, the velocity of object is given by,
${ V }_{ 1 }=\dfrac { { S }_{ 1 } }{ { t }^{ ' } }$
Substituting the values we get,
${ V }_{ 1 }=\cfrac { 10 }{ 2 }$
$\Rightarrow { V }_{ 1 }=5{ m }/{ s }$
Similarly, Displacement from t= 2 s to t= 4 s is given by,
${S}_{21}= {x}_{3}- {x}_{2}$
Substituting the values in above equation we get,
${S}_{2}= 48.5 – 18.5$
$\Rightarrow {S}_{1}= 30 m$
Time taken by the object to move from ${x}_{2}$ to ${x}_{3}$ is given by,
${ t }^{ '' }=4-2$
$\Rightarrow { t }^{ '' }=2 s$
Therefore, the average velocity of object is given by,
${ V }_{ 2 }=\dfrac { { S }_{ 2 } }{ { t }^{ '' } }$
Substituting the values we get,
${ V }_{ 2 }=\cfrac { 30 }{ 2 }$
$\Rightarrow { V }_{ 2 }=15{ m }/{ s }$
Hence, the velocity of the object at t= 0 s and t= 2.0 s is 5 m/s. Its average velocity at t= 2.0 s and t= 4.0 s is 15 m/s.

Note:
Students should understand that the average velocity is not the same as the velocity. Velocity is defined as the displacement in a particular point of time. While the average velocity is the total change in the position or displacement for a given interval of time. To solve these types of problems, students should know the basic formulas.