At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of\({\bf{2}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\). (a) How far does she travel in the next 5.00 s? (b) What is her final velocity? (c) Evaluate the result. Does it make sense?

Short Answer

Expert verified

(a) The distance travelled by racer is\({\bf{20}}\;{\bf{m}}\).

(b) The final velocity of racer is\({\bf{1}}\;{\bf{m/s}}\).

(c) The result does not make any sense.

Step by step solution

01

Determination of distance travelled by racer

(a)

Given Data:

The initial velocity of racer is\(u = 9\;{\rm{m}}/{\rm{s}}\)

The deceleration of racer is\(a = 2\;{\rm{m}}/{{\rm{s}}^2}\)

The time for distance travelled is\(t = 5\;{\rm{s}}\)

The distance travelled by the racer can be found by using second equation of motion with deceleration.

The distance travelled by the racer is given as:

\(d = ut - \frac{1}{2}a{t^2}\)

Here,\(a\)is the deceleration of racer.

Substitute all the values in the above equation.

\(\begin{array}{l}d = \left( {9\;{\rm{m}}/{\rm{s}}} \right)\left( {5\;{\rm{s}}} \right) - \frac{1}{2}\left( {2\;{\rm{m}}/{{\rm{s}}^2}} \right){\left( {5\;{\rm{s}}} \right)^2}\\d = 20\;{\rm{m}}\end{array}\)

Therefore, the distance travelled by racer is \(20\;{\rm{m}}\).

02

Determination of final velocity of racer

(b)

The final velocity of racer is given as:

\({v^2} = {u^2} - 2ad\)

Here,\(v\)is the final velocity of racer.

Substitute all the values in the above equation.

\(\begin{array}{c}{v^2} = {\left( {9\;{\rm{m}}/{\rm{s}}} \right)^2} - 2\left( {2\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {20\;{\rm{m}}} \right)\\v = 1\;{\rm{m}}/{\rm{s}}\end{array}\)

Therefore, the final velocity of racer is \(1\;{\rm{m}}/{\rm{s}}\).

03

Evaluation of result

(c)

The time for complete stop of racer with deceleration is given as:

\(V = u - aT\)

Here,\(V\)is the velocity of racer at stop and its value is zero,\(T\)is the duration necessary for stop.

Substitute all the values in the above equation.

\(\begin{array}{l}0 = 9\;{\rm{m}}/{\rm{s}} - \left( {2\;{\rm{m}}/{{\rm{s}}^2}} \right)T\\T = 4.5\;{\rm{s}}\end{array}\)

The duration for complete stop of racer is less than duration for distance travelled by racer so this contradiction in time does not make any sense for the motion of racer.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A graph of v(t) is shown for a world-class track sprinter in a\({\bf{100}} - {\bf{m}}\)race. (See Figure 2.67). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at\(t = {\rm{ }}{\bf{5}}{\rm{ }}{\bf{s}}\)? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?

Consider a grey squirrel falling out of a tree to the ground.

(a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel’s velocity just before hitting the ground, assuming it fell from a height of 3.0 m.

(b) If the squirrel stops in a distance of 2.0 cmthrough bending its limbs, compare its deceleration with that of the airman in the previous problem.

Give an example (but not one from the text) of a device used to measure time and identify what change in that device indicates a change in time.

Land west of the San Andreas fault in southern California is moving at an average velocity of about 6cm/y northwest relative to land east of the fault. Los Angeles is west of the fault and may thus someday be at the same latitude as San Francisco, which is east of the fault. How far in the future will this occur if the displacement to be made is590km northwest, assuming the motion remains constant?

Suppose you throw a rock nearly straight up at a coconut in a palm tree, and the rock misses on the way up but hits the coconut on the way down. Neglecting air resistance, how does the speed of the rock when it hits the coconut on the way down compare with what it would have been if it had hit the coconut on the way up? Is it more likely to dislodge the coconut on the way up or down? Explain.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free