Physics Fundamentals
Physics Fundamentals
2nd Edition
ISBN: 9780971313453
Author: Vincent P. Coletta
Publisher: PHYSICS CURRICULUM+INSTRUCT.INC.
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Chapter 1, Problem 11P
To determine

The total distance travelled and average speed.

Expert Solution & Answer
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Explanation of Solution

Given:

The time interval of motion is 1.00min .

The following speeds are 0.400km/min , 0.240km/min , 0.160km/min , 0.160km/min and 0.320km/min .

Formula Used:

Speed is the distance covered in unit time by an object.

Write the expression for speed of an object.

  v=dt

Here, d is the distance covered, t is the required time v is the speed of object.

Rearrange the above expression in terms of d .

  d=vt ...... (I)

Write the expression for average speed.

  vavg=DT ...... (II)

Here, vavg is the average speed, D is the total distance travelled and T is the total time.

Calculation:

Substitute 0.400km/min for v and 1.00min for t in equation (I).

  d1=(0.400km/min)(1.00min)=0.400km

Substitute 0.240km/min for v and 1.00min for t in equation (I).

  d2=(0.240km/min)(1.00min)=0.240km

Substitute 0.160km/min for v and 1.00min for t in equation (I).

  d3=(0.160km/min)(1.00min)=0.160km

Substitute 0.160km/min for v and 1.00min for t in equation (I).

  d4=(0.160km/min)(1.00min)=0.160km

Substitute 0.320km/min for v and 1.00min for t in equation (I).

  d5=(0.320km/min)(1.00min)=0.320km

Total distance travelled.

  D=0.400km+0.240km+0.160km+0.160km+0.320km=1.28km

Total time taken.

  T=1.00min+1.00min+1.00min+1.00min+1.00min=5.00min

Substitute 1.28km for D and 5.00min for T in equation (II).

  vavg=1.28km5.00min=0.256km/min

Conclusion:

Thus, the total distance travelled is 1.28km and the average speed is 0.256km/min .

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