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?

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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

For part a:

To calculate the speed when the ball hits the ground, you can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

In this case, the initial velocity (u) is 2.8 m/s, the acceleration (a) is -9.8 m/s^2 (negative because it is acting downward), and the displacement (s) is the distance from the window to the ground, which is 3.6 m.

Using the equation, we can rearrange it to solve for the final velocity (v):

v^2 = u^2 + 2as
v^2 = (2.8 m/s)^2 + 2(-9.8 m/s^2)(-3.6 m)
v^2 = 7.84 m^2/s^2 + 70.56 m^2/s^2
v^2 = 78.4 m^2/s^2
v = sqrt(78.4 m^2/s^2)
v ≈ 8.85 m/s

Therefore, the ball's speed when it hits the ground is approximately 8.85 m/s.

For part b:

To determine how long after the first ball is thrown the second ball should be dropped so that both balls hit the ground at the same time, you need to consider the time it takes for the first ball to reach the ground.

You can calculate the time it takes for the first ball to hit the ground using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is time.

In this case, the initial velocity (u) is 2.8 m/s, the acceleration (a) is -9.8 m/s^2 (negative because it is acting downward), and the final velocity (v) is 0 m/s (since the ball hits the ground).

Using the equation, we can rearrange it to solve for time (t):

v = u + at
0 m/s = 2.8 m/s + (-9.8 m/s^2)t
-2.8 m/s = -9.8 m/s^2t
t = -2.8 m/s / -9.8 m/s^2
t ≈ 0.29 s

Therefore, the first ball takes approximately 0.29 s to hit the ground.

To ensure that both balls hit the ground at the same time, the second ball should be dropped 0.29 s after the first ball is thrown.

Please note that in this calculation, we assume no air resistance and neglect any other external factors that may affect the motion of the balls.

For the first question:

c) To find out how far in front of the truck the car is 10.0 seconds later, we need to calculate the distance covered by the car during the constant velocity phase.

The initial velocity of the car was 28 m/s and it maintained this velocity for 10.0 seconds, so we can use the equation d = vt, where d is the distance, v is the velocity, and t is the time.

Substituting in the values, we get d = (28 m/s)(10 s), which gives us 280 meters. Therefore, after 10.0 seconds, the car is 280 meters in front of the truck.

For the second question:

a) To find the ball's speed when it hits the ground, we need to calculate the final velocity.

We can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

Since the ball is thrown upward, the acceleration due to gravity acts in the opposite direction, so we need to use a negative value for acceleration. Let's take upward as the positive direction, so the initial velocity (u) is 2.8 m/s, the acceleration (a) is -9.8 m/s^2, and the displacement (s) is -3.6 m (negative because the ball is moving upward).

Plugging these values into the equation, we get v^2 = (2.8 m/s)^2 + 2(-9.8 m/s^2)(-3.6 m).

Simplifying, we have v^2 = 7.84 m^2/s^2 + 70.56 m^2/s^2.

Adding the terms together, we get v^2 = 78.4 m^2/s^2.

Finally, taking the square root of both sides of the equation to find v, we have v = √(78.4 m^2/s^2). This gives us approximately 8.85 m/s. Therefore, the ball's speed when it hits the ground is approximately 8.85 m/s.

b) To determine how long after the first ball is thrown the second ball should be dropped from the same window so that both balls hit the ground at the same time, we need to find the time of flight for the first ball.

To calculate the time of flight, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Since the final velocity is 0 m/s (when the ball hits the ground), the initial velocity (u) is 2.8 m/s, and the acceleration (a) is -9.8 m/s^2, we can rearrange the equation to solve for time.

0 = 2.8 m/s + (-9.8 m/s^2)t.

Simplifying, we have -9.8 m/s^2t = -2.8 m/s.

Dividing both sides of the equation by -9.8 m/s^2, we get t = (-2.8 m/s) / (-9.8 m/s^2).

Calculating this, we have t ≈ 0.286 seconds.

Therefore, if the second ball is dropped at the same time as the first ball is thrown, both balls will hit the ground at the same time.