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?
$\red{ \texttt{Explanation is also needed}}$
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Answer 1

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

Theory :

• Velocity

The rate of change of displacement of a particle with time is called velocity of the particle.

$\sf\:Velocity=\dfrac{Distance}{Time\:interval}$

In differential form:

$\sf\:Velocity,V=\dfrac{dx}{dt}$

• Average Velocity:

The average velocity of an object is equal to the ratio of the displacement the time interval for the motion take place.

$\sf\:Average\:Velocity=\dfrac{Displacement}{Time\:Taken}$

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

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

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

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

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

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

Displacement

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

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

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

Time taken = 4-2 sec = 2 sec

We know that

$\sf\:Average\:Velocity=\dfrac{Displacement}{Time\:Taken}$

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

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

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

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

Answer 2

Answer:

Given :-

$\mapsto$ The position of an object moving along x-axis is given by a + bt² (where a = 8.5 m, b = 2.5 m/s² and t is measured in second.

To Find :-

$\mapsto$ What is the average velocity between t = 2.0s and t = 4.0s ?

Solution :-

▪️By using differential calculus, velocity is given by :-

$\leadsto$ V = $\dfrac{dx}{dt}$

$\implies$ $\dfrac{d}{dt}$ (a + bt²)

$\implies$ 2bt

By putting the value we get,

$\implies$ V = 2 × 2.5 ms-¹ × t

$\dashrightarrow$ V = 5.0 t ms-¹

At

• t = 0 s
• v = 0 ms-¹

And at

• t = 2.0 s
• v = 10 ms-¹

Now,

$\mapsto$ x(t) = 4.0 s = a + 16b;

$\mapsto$ x(t = 2.0 s) = a + 4b

We know that,

$\green\bigstar$ Average velocity = $\dfrac{Displacement}{Time}$ $\green\bigstar$

▪️ According to the question,

$\implies$ $\sf{V_{av}}$ = $\dfrac{x(t = 4.0 s) - x(t = 2.0 s)}{4.0 s - 2.0 s}$

$\implies$ $\dfrac{a + 16b - a - 4b}{2.0 b}$

$\implies$ $\dfrac{12b}{2.0 s}$ = 6.0 × b

$\implies$ 6.0 × 2.5

$\dashrightarrow$ $\sf{V_{av}}$ = 15ms-¹

$\therefore$ The average velocity between t = 2 sec and 4 sec is $\boxed{\bold{\large{15 ms-¹}}}$

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