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Q17P

Expert-verifiedFound in: Page 33

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**

(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.

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

**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:**

${v}_{avg}=\frac{\u2206x}{\u2206t}$ (i)

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

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,

${\mathrm{v}}_{\mathrm{avg}}=\frac{\u2206\mathrm{x}}{\u2206\mathrm{t}}\phantom{\rule{0ex}{0ex}}=\frac{50.25-21.75}{3.00-2.00}\phantom{\rule{0ex}{0ex}}=28.5\mathrm{cm}/\mathrm{s}$

Average velocity is 28.5 cm/s

Using equation (ii) the instantaneous velocity is

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

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

Instantaneous velocity at *t*=3.00 s,

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

Instantaneous velocity at *t*=2.50 s

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

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

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

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)

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