Question

The position of an object moving along x - axis is given by a + bt² where a = 8.5m b = 2.5 m/s² and t is measured in seconds.
what is it's velocity at t = 0 and t = 2.0 s.
what is the average velocity between t = 2.0 s and t = 4.0 s?

$\sf \bold {Full \ Explanation \Needed}$
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Answer 1

$\huge{\mathcal{\underline{\green{Answer}}}}$

Position is given as x=a+bt²

=8.5+2.5t²

Position at t=2 s, x₂ =8.5+2.5(2)²

=18.5 m

Position at t=4 s, x₁ =8.5+2.5(4)²

=48.5 m

Displacement S= x₂-x₁

=48.5−18.5

=30 m

Time taken t=4−2=2 s

Average velocity Vavg =$\frac{S}{ T }$

=30/2

=15m/s

Answer 2

Given :

• The position of an object moving along x - axis is given by a + bt²
• Where a= 8.5 and b = 2.5

To Find :

1. Velocity of an object at t= 0 and t = 2 sec
2. Average velocity between t= 2 sec and 4 sec.

Solution :

The position of an object moving along x - axis is given by $\sf\:x(t)=a+bt^2$

1) We have to find Velocity of an object at t= 0 and 2 sec.

$\sf\:x(t)=a+bt^2$

Now , Differentiate with respect to t

$\sf\dfrac{dx}{dt}=\dfrac{d(a)}{dt}+\dfrac{d(bt^2)}{dt}$

$\sf\implies\dfrac{dx}{dt}=0+2bt$

$\sf\implies\dfrac{dx}{dt}=2bt$

$\sf\implies\:v=2bt$

At t = 0 Sec

$\sf\:v=2b\times0$

$\sf\implies\:v=0ms^{-1}$

At t = 2 sec

$\implies\sf\:v=2b\times2$

$\sf\implies\:v=4b$

Now put the value of b,

$\sf\implies\:v=4\times2.5$

$\sf\implies\:v=10ms^{-1}$

Therefore , Velocity at t= 0 is 0 m/s and at t = 2 sec Velocity is 10 m/s.

2) We have to find the average velocity between t = 2 sec and 4 sec

$\sf\:x(t)=a+bt^2$

At t = 2 sec,

$\sf\:x(2)=a+b(2)^2$

$\sf\:x(2)=a+4b$

Now put the values of a & b,

$\sf\:x(2)=8.5+4\times2.5$

$\sf\:x(2)=18.5m$

At t = 4 sec

$\sf\:x(4)=a+b(4)^2$

$\sf\:x(4)=a+16b$

Now put the values of a & b,

$\nrightarrow \: \sf\:x(4)=8.5+16\times2.5$

$\nrightarrow\sf\:x(4)=48.5$

Displacement,

$\longrightarrow\sf\triangle\:x=x(4)-x(2)$

$\longrightarrow\sf\triangle\:x=48.5-18.5$

$\longrightarrow \: \sf\triangle\:x=30m$

Time taken = 4-2 sec = 2 sec,

We know that,

$\bf\: \implies \: \underline{\boxed{ \bf{Average\:Velocity=\dfrac{Displacement}{Time\:Taken}}}}$

$\sf\implies\:v=\dfrac{\triangle\:x}{dt}$

$\sf\implies\:v=\dfrac{30}{2}$

$\sf\implies\:v=15ms^{-1}$

Therefore, the average velocity between t= 2 sec and 4 sec is 15 m/s.