posted by Alex on .
A car is traveling at 20m/s when it pulls out to pass a truck that is traveling at only 18m/s. The car accelerates at 2.0m/s^2 for 4.0 s and then maintains this new velocity.
First it asked me to calculate the distance during acceleration so I used d = (v1)(t)+[(a)(t^2)]/2 and got 56 km.
Then it asked me to calculate the car's speed so I used V1+(a)(t) and got 28 m/s.
Now c) If the car was originally 8.0 m behind the truck when it pulled out to pass, how far in front of the truck is the car 10.0 s later?
A ball is thrown vertically upward from a window that is 3.6 m above the ground. The ball's initial speed is 2.8 m/s and the acceleration due to gravity is 9.8 m/s^2.
a) What is the ball's speed when it hits the ground?
b) How long after the first ball is thrown should a second ball be simply dropped from the same window so that both balls hit the ground at the same time?
For part a do I calculate the final velocity first? and then use the final velocity as my initial velocity? Should I also change my acceleration to negative?
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