Average Velocity and Average Speed Problems and Solutions 3

  Problem#1

You are to drive to an interview in another town, at a distance of 300 km on an expressway. The interview is at 11:15 A.M. You plan to drive at 100 km/h, so you leave at 8:00 A.M. to allow some extra time. You drive at that speed for the first 100 km, but then construction work forces you to slow to 40 km/h for 40 km. What would be the least speed needed for the rest of the trip to arrive in time for the interview?

Answer:
The values used in the problem statement make it easy to see that the first part of the trip (at 100 km/h) takes 1 hour, and the second part (at 40 km/h) also takes 1 hour.

Expressed in decimal form, the time left is 1.25 hour, and the distance that remains is 160 km.

Thus, a speed v = (160 km)/(1.25 h) = 128 km/h is needed.

Problem#2
Traffic shock wave. An abrupt slowdown in concentrated traffic can travel as a pulse, termed a shock wave, along the line of cars, either downstream (in the traffic direction) or upstream, or it can be stationary. Figure 1 shows a uniformly spaced line of cars moving at speed v = 25.0 m/s toward a uniformly spaced line of slow cars moving at speed v_s = 5.00 m/s. Assume that each faster car adds length L = 12.0 m (car length plus buffer zone) to the line of slow cars when it joins the line, and assume it slows abruptly at the last instant. (a) For what separation distance d between the faster cars does the shock wave remain stationary? If the separation is twice that amount, what are the (b) speed and (c) direction (upstream or downstream) of the shock wave?

Fig.1

Answer:
(a) Let the fast and the slow cars be separated by a distance d at t = 0. If during the time interval
t = L/v_s = (12.0 m)/(5.0 m/s) = 2.40 s in which the slow car has moved a distance of L =12.0 m ,
the fast car moves a distance of vt = d + L to join the line of slow cars, then the shock wave would remain stationary.

The condition implies a separation of

d = vt − L = (25 m/s)(2.4 s) − 12.0 m = 48.0 m.

(b) Let the initial separation at t = 0 be d = 96.0 m. At a later time t, the slow and the fast cars have traveled x = v_st and the fast car joins the line by moving a distance d + x . From

t = x/v_s = (d + x)/v

we get

x = v_sd/(v – v_s)

x = 5.00 x 96.0 m/(25.0 m/s – 5.00 m/s) = 24.0 m

which in turn gives

t = (24.0 m) /(5.00 m/s) = 4.80 s

Since the rear of the slow-car pack has moved a distance of ∆x = x − L = 24.0 m − 12.0 m =12.0 m downstream, the speed of the rear of the slow-car pack, or equivalently, the speed of the shock wave, is

v_shock = ∆x/t = 12.0 m/4.8 s = 2.5 m/s

(c) Since x > L , the direction of the shock wave is downstream.

Problem#3
You drive on Interstate 10 from San Antonio to Houston, half the time at 55 km/h and the other half at 90 km/h. On the way back you travel half the distance at 55 km/h and the  other half at 90 km/h. What is your average speed (a) from San Antonio to Houston, (b) from Houston back to San Antonio, and (c) for the entire trip? (d) What is your average velocity for the entire trip? (e) Sketch x versus t for (a), assuming the motion is all in the positive x direction. Indicate how the average velocity can be found on the sketch.

Answer:
(a) Denoting the travel time and distance from San Antonio to Houston as T and D, respectively, the average speed is

S_avg.1 = D/T  = [(55 km/h)(T/2) + (90 km/h)(T/2)]/T = 72.5 km/h ≈ 73 km/h

 (b) Using the fact that time = distance/speed while the speed is constant, we find

S_avg2 = D/T = D/[(½D/55 km/h) + (½D/90 km/h)]

S_avg2 = 68.3 km/h ≈ 68 km/h

(c) The total distance traveled (2D) must not be confused with the net displacement (zero). We obtain for the two-way trip

S_avg = 2D/[(D/72.5 km/h) + (D/68.3 km/h)] = 70 km/h

(d) Since the net displacement vanishes, the average velocity for the trip in its entirety is zero.

(e) In asking for a sketch, the problem is allowing the student to arbitrarily set the distance D (the intent is not to make the student go to an atlas to look it up); the student can just as easily arbitrarily set T instead of D, as will be clear in the following discussion. We briefly describe the graph (with kilometers-per-hour understood for the slopes): two contiguous line segments, the first having a slope of 55 and connecting the origin to (t₁, x₁) = (T/2, 55T/2) and the second having a slope of 90 and connecting (t₁, x₁) to (T, D) where D = (55 + 90)T/2. The average velocity, from the graphical point of view (Fig.2), is the slope of a line drawn from the origin to (T, D). The graph (not drawn to scale) is depicted below:

Fig.2

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