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Found in: Page 33

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# The position of a particle moving along x axis is given in cm by ${\mathbf{x}}{\mathbf{=}}{\mathbf{9}}{\mathbf{.}}{\mathbf{75}}{\mathbf{+}}{\mathbf{1}}{\mathbf{.}}{\mathbf{50}}{{\mathbf{t}}}^{{\mathbf{3}}}$, where t is in seconds. Calculate (a) the average velocity during time interval t=2 s to t=3 s; (b) the instantaneous velocity at t=2 s; (c) the instantaneous velocity at t=3s; (d) the instantaneous velocity at t=2.5 s; (e) the instantaneous velocity when the particle is midway between its positions at t=2 s and t=3 s (f) Graph x vs. t and indicate your answers graphically.

(a) The average velocity during t=2.00 s to t=3.00 s is 28.5 cm/s.

(b) Instantaneous velocity at t=2.00 s is 18.0 cm/s

(c) Instantaneous velocity at t=3.00 is 40.5 cm/s.

(d) Instantaneous velocity at t=2.50 s is 28.1 cm/s.

(e) Instantaneous velocity when particle is midway between t=2.00s to is 30.3 cm/s.

See the step by step solution

## Step 1: Given information

$\mathrm{x}=9.75+1.50{\mathrm{t}}^{3}$

## Step 2: To understand the concept of average velocity

The problem deals with the average velocity which is the displacement over the time. Also, it involves the instantaneous velocity of an object. It is the limit of an average velocity as the elapsed time approaches zero. Using a standard equation the average velocity in terms of displacement and time can be found. Further, instantaneous velocity can be found by taking derivative of equation of position.

Formula:

${v}_{avg}=\frac{∆x}{∆t}$ (i)

${\mathrm{v}}_{\mathrm{instantaneous}}=\frac{\mathrm{dx}}{\mathrm{dt}}$ (ii)

## Step 3: (a) Calculate the average velocity during time interval t=2. s to t=3 s.

It is given that,

$\mathrm{x}=9.75+1.50{\mathrm{t}}^{3}$

By plugging the values of t in the above equation we get,

${\mathrm{x}}^{2}=9.75+1.50{\left(2.00\right)}^{3}\phantom{\rule{0ex}{0ex}}=21.75\mathrm{cm}\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{3}=9.75+1.50{\left(3.00\right)}^{3}\phantom{\rule{0ex}{0ex}}=50.25\mathrm{cm}$

Using equation (i) the average velocity is,

${\mathrm{v}}_{\mathrm{avg}}=\frac{∆\mathrm{x}}{∆\mathrm{t}}\phantom{\rule{0ex}{0ex}}=\frac{50.25-21.75}{3.00-2.00}\phantom{\rule{0ex}{0ex}}=28.5\mathrm{cm}/\mathrm{s}$

Average velocity is 28.5 cm/s

## Step 4: (b) Calculate the instantaneous velocity at t=2 s

Using equation (ii) the instantaneous velocity is

$v=\frac{dx}{dt}\phantom{\rule{0ex}{0ex}}=4.5{t}^{2}$

At $\mathrm{t}=2.00\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{v}=4.5{\left(2.00\right)}^{2}\phantom{\rule{0ex}{0ex}}=18.0\mathrm{cm}/\mathrm{s}$

## Step 5: (c) Calculate the instantaneous velocity at t=3 s

Instantaneous velocity at t=3.00 s,

$\mathrm{v}=4.5{\left(3.00\right)}^{2}\phantom{\rule{0ex}{0ex}}=40.5\mathrm{cm}/\mathrm{s}$

## Step 6: (d) Calculate the instantaneous velocity at t=2.5 s

Instantaneous velocity at t=2.50 s

$\mathrm{v}=4.5{\left(2.50\right)}^{2}\phantom{\rule{0ex}{0ex}}=28.1\mathrm{cm}/\mathrm{s}$

## Step 7: (e) the instantaneous velocity when the particle is midway between its positions at t=2 s and t=3 s

Trying to find time of particle at midway between ${x}_{2}$ and ${x}_{3}$, we get,

${\mathrm{x}}_{\mathrm{m}}=\frac{50.25+21.75}{2}\phantom{\rule{0ex}{0ex}}=36\mathrm{cm}\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{\mathrm{m}}=9.75+1.50{\mathrm{t}}_{\mathrm{m}}^{3}\phantom{\rule{0ex}{0ex}}36=9.75+1.50{\mathrm{t}}_{\mathrm{m}}^{3}\phantom{\rule{0ex}{0ex}}{\mathrm{t}}_{\mathrm{m}}=2.596\mathrm{s}$

Time at midway is 2.596 s

Thus, instantaneous speed at this time

$\mathrm{v}=4.5\left(2.596\right)2\phantom{\rule{0ex}{0ex}}=30.3\mathrm{cm}/\mathrm{s}$

## Step 8: (f) Graph x vs t and indicate the answers

Straight line slope between t=2s to 3s gives answer of part (a) and slope of tangent at the required points would give the answers of (b), (c), (d), (e)