Physics problem.. HELP!!!

Question

Car A is traveling on a highway at a constant speed (VA)o= 100km/h and is 120m from the entrance of an access ramp when car B enters the acceleration lane at that point at a speed (VB)o =25 km/hr. Car B accelerates uniformly and enters the main traffic lane after traveling 70 m in 5s. It then continues to accelerate at the same rate until it reaches a speed of 100km/hr, which it then maintains. Determine the final distance between the two cars.

Answer 1

Hi Dianne-

Somu does not "get it" for whatever reason.  That's OK (everyone draws a blank from time to time).

The cars are clearly traveling in the same direction...

If you don't get any step, just ask, and I'll elaborate.

We need our motion equations, like vf^2 = vo^2 + 2*a*x and vf=vo+a*t etc. (If you need me to list those "Big 4" for you, please tell me and I'll do it.

We also need Dist=rate * time for the other car (at constant velocity).

Car B travels for 5 seconds and covers 70m (from a vo=25 km/hr = 6.94 m/sec).  The accel = 2.82 m/s^2 and vf = 21 m/s

Using 21m/s as vo and 2.82 m/s^2 as accel and vf = 100km/hr = 27.8 m/s, we get a time of 2.41 sec and a distance traveled of 58.8 m

There are these two pieces to finding the distance traveled by car B, so...Car B has thus wound up at 100 km/hr in 7.41 seconds and covered a distance of 70 + 58.8 = 128.8 m

Car A will travel at 27.8 m/s for 7.41 seconds and cover 206 m (but it started out 120m behind car B)

(Note that when Car B entered traffic after 5 seconds, Car A had gone 139 meters or was 19 meters past the entrance.)

Car B is in front of Car A by 128.8m - 86 m = 42.8 meters

(I got the 86 meters from 206m-120m).

-Fred

Check my numbers, but the method is fine.