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HI Dr. Johnson!

I'm a undergraduate and currently taking a general physics course. I have never been introduced to physics before.

My question is concerning motion in a straight line and is the following:

a red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis. At time t = 0, the red car is at x(r)= 0 and the green car is at x(g)= 220 m. If the red car has a constant velocity of 20 km/h, the cars pass each other at x = 44.5 m, and if it has a constant velocity of 40 km/h, they pass each other at x = 76.6 m. What are (a) the initial velocity and (b) the constant acceleration of the green car?

Hello Edward,

The time will be the same for both, so call the time to meet Tm. Let the distance traveled by the red and green cars be Dr and Dg respectively. We then know that

Dr + Dg = 220 m.

Let's convert 20 km/h and 40 km/h to m/s.

20 km/h * (1000 m/1 km) * (1 h/3600 s) = 5.56 m/s

40 km/h then would be 11.1 m/s

If the red car's constant velocity is 5.56 m/s, then

Tm = Dr/red velocity = 44.5 m / (5.56 m/s) = 8 s

Applying the kinematic formula

x = Vi*t + (1/2)*a*t^2

to the green car, we have

(220 m - 44.5 m) = Vi*8 s + (1/2)*a*(8 s)^2

Note: these units are valid since we have meters on the left and on the right we have velocity times time and acceleration times time squared. The units are cluttering up the equation, let's have faith that Vi will be in m/s and a will be in m/s^2 and drop the units.

175.5 = 8.0*Vi + 32*a ........ Eq1

If the red car's constant velocity is 11.1 m/s, then

Tm = 76.6 m / (11.1 m/s) = 6.9 s

Applying the kinematic formula

x = Vi*t + (1/2)*a*t^2

to the green car, we have

(220 m - 76.6 m) = Vi*6.9 s + (1/2)*a*(6.9 s)^2

143.4 = 6.0 Vi + 23.8*a ........ Eq2

Let's solve Eq1 and Eq2 for Vi and then set the expressions for Vi equal to each other.

175.5 = 8*Vi + 32*a ........ Eq1

Vi = (175.5 - 32*a)/8

143.4 = 6*Vi + 23.8*a ........ Eq2

Vi = (143.4 - 23.8*a)/6

Vi = Vi so

(175.5 - 32*a)/8 = (143.4 - 23.8*a)/6

(175.5 - 32*a) = 8*(143.4 - 23.8*a)/6

175.5 - 32*a = 191.2 - 31.7*a

0.3*a = -15.7

a = -15.7/0.3 = -52.3 m/s^2

Substituting that for the a in Eq1 or Eq2 should allow you to find Vi.

I did that and find that neither of my results are reasonable for any car I've driven. Perhaps your instructor has a devious side or maybe there is an error in my arithmetic (possibly even in my algebra). Even if the results were reasonable, I would suggest that you go through these steps critically.

I hope this helps,

Steve

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Comment | Thank you for the answer. I appreciate the step by step structure. Sorry for the late reply and thanks again! |

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I would be delighted to help with questions up through the first year of college Physics. Particularly Electricity, Electronics and Newtonian Mechanics (motion, acceleration etc.). I decline questions on relativity and Atomic Physics. I also could discuss the Space Shuttle and space flight in general.

I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.**Education/Credentials**

BS Physics, North Dakota State University

MS Electrical Engineering, North Dakota State University