Chapter 02, Problem 034 In the figure here, a red car and a green car move towar

Question

Chapter 02, Problem 034 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 atxg 224 m. If the red car has a constant velocity of 21.0 km/h, the cars pass each other at x 44.2 m. On the other hand, if the red car has a constant velocity of 42.0 km/h, they pass each other at x = 76.8 m, what are (a) the initial velocity and (b) the (constant) acceleration of the green car? Include the signs. Green car car Unit (a) Number (b) Number Click if you would like to Show Work for this question: Unit Open Show work

Explanation / Answer

first of all we must convert the speeds from kmph to m/s

1 kmph = (5 / 18) m/s

21 kmph = (21) (5 / 18)

= 5.933 m/s

so the velocity of the red car will be v1 = 5.833 m/s and

v2 = (42) (5 / 18)

= 11.66 m/s

let d be the distance between the two cars at t = 0 which is d = xg = 224 m and

x1 = 44.2 m

x2 = 76.8 m

here we shall make use of the equation of motion

s = ut + (1 / 2) a t2

where s is the displacement, u is the initial velocity a is the acceleration and t is the time taken

in the first case the displaement is d-x1 and the in the second case the displaement is d-x2

in the first case the time taken will be t1 = x1/v1 = 7.57 s

since it is constant velocity

in the second case the time taken will be t2 = x2/v2 = 6.58 s

then the two equatuions of mition will be

d - x1 = u t1 + (1 / 2) a t12 and

d - x2 = u t2 + (1 / 2) a t22

solving for the values of intial velocity and acceleration we get

224 - 44.2 = u (7.57) + (1 / 2) a (7.57)2 and

224 - 76.8 = u (6.58) + (1 / 2) a (6.58)2

acceleration as a = - 2.79 m / s2

initial velocity will be u = - 13.67 m/s

since both the values are negative hence the motion is along negative x-direction

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