MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months)
MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months)
5th Edition
ISBN: 9781305581159
Author: Nicholas J. Garber; Lester A. Hoel
Publisher: Cengage Learning US
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Chapter 3, Problem 9P
To determine

(a)

The distance traveled by the vehicle when the acceleration is 45 ft/sec.

Expert Solution
Check Mark

Answer to Problem 9P

  x=443ft

Explanation of Solution

Given information:

  dudt=3.60.06uα=3.6β=0.06

Initial vehicle speed, u0=30ft/sec

Final vehicle speed, ut=45ft/sec

Concept used:

The time required by the vehicle upon acceleration is calculated by using the relation,

  βt=ln[( αβ u t )( αβ u 0 )]t=(1β)ln[( αβ u t )( αβ u 0 )]

The distance travelled at any time is calculated by using the formula,

  x=(αβ)tαβ2(1eβt)+u0β(1eβt)

Calculation:

The time taken by the vehicle is calculated as,

  t=(1β)ln[( αβ u t )( αβ u 0 )]=(1 0.06)ln[3.60.06( 45)3.60.06( 30)]=11.55sec

  x=(αβ)tαβ2(1e βt)+u0β(1e βt)=( 3.6 0.06)11.553.6 ( 0.06 )2(1e ( 0.06×11.55 ))+300.06(1e ( 0.06×11.55 ))=693499.93+249.96=443ft

Conclusion:

The distance traveled when the vehicle has accelerated to 45 ft/sec is 443ft.

To determine

(b)

The time taken by the vehicle to attain a speed of 45 ft/sec.

Expert Solution
Check Mark

Answer to Problem 9P

  t=11.55sec

Explanation of Solution

Given information:

  dudt=3.60.06uα=3.6β=0.06

Initial vehicle speed, u0=30ft/sec

Final vehicle speed, ut=45ft/sec

Concept used:

The time required by the vehicle after accelerated is calculated by using the relation,

  βt=ln[( αβ u t )( αβ u 0 )]t=(1β)ln[( αβ u t )( αβ u 0 )]

Calculation:

  t=(1β)ln[( αβ u t )( αβ u 0 )]=(1 0.06)ln[3.60.06( 45)3.60.06( 30)]=11.55sec

Conclusion:

The time for vehicle to attain speed of 45 ft/sec is 11.55sec.

To determine

(c)

The acceleration of the vehicle after 4 seconds.

Expert Solution
Check Mark

Answer to Problem 9P

  a=1.42fts2

Explanation of Solution

Given information:

  dudt=3.60.06uα=3.6β=0.06

Initial vehicle speed, u0=30ft/sec

Final vehicle speed, ut=45ft/sec

Time taken, t=4sec

Concept used:

The initial velocity is calculated and then the acceleration is calculated.

  ut=αβ(1eβt)+u0eβt

  a=dudt

Calculation:

  ut=αβ(1e βt)+u0eβt=3.60.06(1e ( 0.06×4 ))+30(e ( 0.06×4 ))=3.60.06(10.78663)+30(0.78663)=36.40ft/sec

  a=dudt=3.60.06u=3.60.06(36.40)=1.42fts2

Conclusion:

The acceleration after 4 seconds is 1.42fts2.

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