A car travels along a straight line at a constant speed of 59.0 mi/h for a dista

Question

A car travels along a straight line at a constant speed of 59.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average velocity for the entire trip is 25.0 mi/h. (a) What is the constant speed with which the car moved during the second distance d? mi/h (b) Suppose the second distance d were traveled in the opposite direction; you forgot something and had to return home at the same constant speed as found in part (a). What is the average velocity for this trip? mi/h (c) What is the average speed for this new trip? mi/h

Explanation / Answer

let,

initially,

speed of the car, v1=59 mi/hr

distance travelled is d1=d

time taken, t1=d1/v1

and then,

again speed of the car is v2

distance travelled is d2=d

time taken, t2=d2/v2

avd velocity, vavg=25 mi/h

a)

use,

total time= t1+t2

2*d/25=d1/59 + d2/v2

2*d/25=d/59 + d/v2

2/25=1/59 + 1/v2

====>

v2=15.86 mi/hr ----------> is answer

b)

since the distance travelled is same,

avergage velocity, Vavg=25 mi/hr

c)

average speed, vavg=25 mi/hr

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