In the figure here, a red car and a green car move toward each other in adjacent

Question

In the figure here, a red car and a green car move toward each other in adjacent lanes and parallel to an x axis. At time t = 0, the red car is at x_ = 0 and the green car is at x_g = 224 m. If the red car has a constant velocity of 22.0 km/h, the cars pass each other at x = 43.6 m. On the other hand, if the red car has a constant velocity of 44.0 km/h, they pass each other at x = 76.5 m, what are (a) the initial velocity and (b) the (constant) acceleration of the green car? Include the signs. (a) Number _______ unit _______ (b) Number ______ unit ______ Click if you would like to Show Work for this question: Open Show Work

Explanation / Answer

Solution:

Give that the red car starts at x=0 and covers 43.6 m when its velocity is equal to 22.0 km/h. = 6.11 m/s

and the red car starts at x=0 and covers 76.5 m when its velocity is equal to 44.0 km/h. = 12.22 m/s

The time that passes before the cars meet will be

v1=d1 / t1

=> t1 = d1 / v1

=> t1 = 43.6 / 6.11 = 7.14 sec

and

t2 = 76.5 / 12.22 = 6.26 sec

Now , for green car. In the first case, We have

43.5 = 224 + u*t1+ 0.5 *a * t 1^2 .............(1).

Similarly for the second case, We have

76.5 = 224 + u * t2 + 0.5*a*t2^2 ...........(2)

Now , Solving first and second equation.

180.5 = 7.14*u + a*25.49

147.5 = 6.26*u + a*19.59

a = 180.5 7.14*u / 25.49 = 7.08 0.28 * u ...............(3)

Putting value of a in second

147.5 = 6.26* u + 19.59 * (7.08 0.28 * u)

147.5 = 6.26*u 138.7 5.48 * u

u = 11.28 m/s

so, using equation 3

a = 7.08 0.28 *(11.28)= 3.92 m/s2

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