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 xr = 0 and the green car is at xg = 224 m. If the red car has a constant velocity of 25.0 km/h, the cars pass each other at x = 44.8 m. On the other hand, if the red car has a constant velocity of 50.0 km/h, they pass each other at x = 76.7 m. What are (a) the initial velocity and (b) the (constant) acceleration of the green car? Include the signs.

(a) Number Unit This answer has no unitskm/hmkgsm/s^2NJWN/mkg•m/sPahgm/s^3times (b) Number Unit This answer has no unitskm/hmkgsm/s^2NJWN/mkg•m/sPahgm/s^3times

Explanation / Answer

here,

initial position of the red car , xr = 0

initial position of the green car , xg = 224 m

initial for case 1

constant velocity of red car , v1 = 25 km/h

v1 = 6.94 m/s

distance travelled by red car , dr1 = 44.8 m

time taken , t1 = dr1/t1

t1 = 44.8 / 6.94

t1 = 6.45 s

for red car

let the initial velocity be u and the accelration be a

the distance travelled , dg1 = 224 - 44.8

dg1 = 179.2 m

using seccon equation of motion

s = ut + 0.5 * at^2

179.2 = u * 6.45 + 0.5 * a* 6.45^2 ......(1)

then

constant velocity of red car , v2 = 50 km/h

v2 = 13.88 m/s

distance travelled by red car , dr2 = 76.7 m

time taken , t2 = dr2/t2

t2 = 76.7 / 13.88

t2 = 5.53 s

for red car

the distance travelled , dg2 = 224 - 76.7

dg2 = 147.3 m

using seccon equation of motion

s = ut + 0.5 * at^2

147.3 = u * 5.53 + 0.5 * a* 5.53^2 ......(2)

solving equation (1) and (2)

u = 19.74 m/s

a = 2.49 m/s^2

(a)

the initial velocity is 19.74 m/s

(b)

the constant acccelration of the green car is 2.49 m/s^2

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