Question

When the drivers pass each other, the driver of the red car is to toss a package of contraband to the other driver. To catch the toss on video, one of the cameras must be set up near the point where the toss will be made. When "action" is yelled out on the set, the initial separation between the drivers will be 200 m, the red car will accelerate from rest at a constant 6.12 m/s2, starting at xr = 0, and the green car will already be in motion with a constant speed of 60 km/h at xg = 200 m. How far from the initial position of the red car will the toss occur?

Answer 1

Answer:

D=387.28m

Explanation:

At the moment where the toss is made $X_R = X_G$, so we need both equations:

For the red car:

$X_R=\frac{a_R*t^2}{2}$   With initial speed of 0 and acceleration of 6.12m/s^2.

For the green car:

$X_G=Xo + V_G*t$   With $V_G = 60km/h*\frac{1000m}{1km} * \frac{1h}{3600s} = 16.66m/s$   and Xo = 200m

Since both positions will be the same:

$\frac{a_R*t^2}{2}=Xo+V_G*t$   Solving for t:

t1 = -5.8s  and   t1 =11.25s

Replacing t = 11.25 on either equation to find the displacement:

$D = X_R = \frac{a_R*t^2}{2} = 387.28m$