Calculating Minimum Deceleration and Meeting Heights in Kinematics

[~SPaRtaN~]

Homework Statement

(1) A speed trap is set up with two pressure activated strips placed across a highway, 116 meters apart. A car is speeding along at 29.6 meters per second, while the speed limit is only 19.4 meters per second. At the instant the car activates the first strip, the driver begins slowing down. What minimum deceleration (magnitude only) is needed so that the driver's average speed is within the limit by the time the car crosses the second strip?

(2)Suppose you throw a stone straight up with an initial velocity of 29.5 m/s and, 4.5 s later you throw a second stone straight up with the same initial velocity. The first stone going down will meet the second stone going up. At what height above the point of release do the two stones meet

Homework Equations

v(final) = v(initial) + at
x(final) = x(initial) + v(initial)*t + 1/2 at^2
x(final) - x(initial) = 1/2 [v(final) - v(initial)]*t
2a[x(final) - x(initial)] = v(final)^2 - v(initial)^2

The Attempt at a Solution

Couldn't get the right answer, tried 2 times

 

[edziura]

You might begin with these two equations:

d = \frac{1}{2} a t²

where a and t are unknown, and

v_avg = \frac{vfinal + vinitial}{2}

where vfinal is unknown but can be written in terms of vinitial, a and t.

 

[~SPaRtaN~]

^ i tried but lost somewhere...its too confusing for me

 

[~SPaRtaN~]

anybody to help on these problems...just 1 day left