How to solve this problem #11 road at an extremely high rate of speed may be about as safe as riding a rhino. You continue toaccelerate at 1.3m/s' until you reach a speed of 32 m/s. How long did it take you to reach thisspeed? How far have you gone? You see a very sharp turn 27m ahead and realize that you needto stop before you reach the turn or else you will go off the road and over the cliff and your

Answered: road at an extremely high rate of speed…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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How to solve this problem #11

road at an extremely high rate of speed may be about as safe as riding a rhino. You continue to
accelerate at 1.3m/s' until you reach a speed of 32 m/s. How long did it take you to reach this
speed? How far have you gone? You see a very sharp turn 27m ahead and realize that you need
to stop before you reach the turn or else you will go off the road and over the cliff and your

Expert Solution

Step 1

The bike starts from rest .Hence the initial velocity u=0 m/s

The acceleration of the bike $a = 1.3 m / s^{2}$

The velocity reached with this velocity v=32 m/s

The time taken 't' to reach the velocity 32 m/s is calculated as follows

$v = u + at t = \frac{v - u}{a} t = \frac{(32 - 0) m / s}{1.3 m / s^{2}} t = 24.62 s$

The distance moved ' $∆ x$ ' in this time is calculated as follows

$v^{2} = u^{2} + 2 a (∆ x) ∆ x = \frac{v^{2} - u^{2}}{2 a} ∆ x = \frac{{(32 m / s)}^{2} - 0}{2 (1.3 m / s^{2})} ∆ x = 393.85 m$ 393.85 m

Step 2

The speed of the bike when the sharp turn is seen $u_{1} = 32 m / s$

The distance moved before the bike is stopped $∆ x_{1} = 27 m$

The final velocity of the bike when it is stopped $v_{1} = 0 m / s$

The acceleration of the bike is

${v_{1}}^{2} = {u_{1}}^{2} + 2 a_{1} ∆ x_{1} 0 = {(32 m / s)}^{2} + 2 a (27 m) a_{1} = \frac{- {(32 m / s)}^{2}}{2 (27 m)} a_{1} = - 18.96 m / s^{2}$

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