Question

The coordinate of a body moving along the X-axis is given by x = 15t2

, where x is in cm and t is

in sec. compute average velocity of the body over the time interval from (a) 3 to 3.1 sec (b) 3 to

3.001 sec (c) 3 to 3.00001 sec (d) what is its instantaneous velocity exactly at 3 sec? ​

Answer 1

Given:

Equation of motion of a object is

$x = 15 {t}^{2}$

To find:

1. Average velocity in the time interval :

• 3 sec to 3.1 sec
• 3 sec to 3.001 sec
• 3 sec to 3.00001 sec

2. Instantaneous Velocity at 3 sec

Calculation:

In time 3 - 3.1 sec:

$v \: avg. = \dfrac{\Delta x}{\Delta t}$

$= > v \: avg. = \dfrac{15 \{ {(3.1)}^{2} - {3}^{2} \}}{3.1 - 3}$

$= > v \: avg. = \dfrac{15 \{ 9.61 - 9 \}}{0.1}$

$= > v \: avg. = \dfrac{15 \{ 0.61\}}{0.1}$

$= > v \: avg. = 91.5 \: cm {s}^{ - 1}$

In time 3 - 3.001 sec:

$v \: avg. = \dfrac{\Delta x}{\Delta t}$

$= > v \: avg. = \dfrac{15 \{ {(3.001)}^{2} - {3}^{2} \}}{3.001 - 3}$

$= > v \: avg. = \dfrac{15 \{ 9.006001 - 9 \}}{0.001}$

$= > v \: avg. = \dfrac{15 \{ 0.006001\}}{0.001}$

$= > v \: avg. = 90.015 \: cm {s}^{ - 1}$

In time 3 - 3.00001 sec:

$v \: avg. = \dfrac{\Delta x}{\Delta t}$

$= > v \: avg. = \dfrac{15 \{ {(3.00001)}^{2} - {3}^{2} \}}{3.00001 - 3}$

$= > v \: avg. = \dfrac{15 \{ 9.0000600001 - 9 \}}{0.00001}$

$= > v \: avg. = \dfrac{15 \{ 0.0000600001\}}{0.00001}$

$= > v \: avg. = 90.00015 \: cm {s}^{ - 1}$

Instantaneous Velocity at 3 sec:

$v = \dfrac{d x}{d t} = 30t$

Putting value of t :

$= > v = 30 \times 3$

$= > v = 90 \: cm {s}^{ - 1}$