Posted by Alex on Monday, October 29, 2007 at 5:49am.
For the accelerating car at the end of 4 s, I get
d = 20 m/s*4s + (1/2)*2m/s^2*16 s^2
= 96 m, not 56 km
The speed at the end of 4s is correct.
The distance travelled by the first car during this interval is 18*4 = 72 m.
If the first car started 8 m behind, it will be 96 - 72 - 8 = 16 m ahead after 4 seconds. For the next 6 sec, up to t=10, the car separation distance will increase at 10 m/s rate, making the total separation 16 + 60 = 76 m.
For you second problem, part (a), you can use conservation of energy to get the velocity at impact.
(1/2)M V2^2 = (1/2) M V1^2 + MgH
V1 = 2.8 m/s and H = 3.6 m. M cancels out. For part (b), calculate the time it takes the upward-thrown ball to reach the ground. Use a negative value of -g for the acceleration, and a positive intial velocity. The second ball will reach the ground sqrt(2H/g) = 0.86 s after it is released
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