Problem #1)

Assume that you are driving a car along a level, straight road at 100 km/hr (you're on a trip in Canada). You see a red light and put on the brakes. It takes you 0.5 seconds from the time you see the light until your foot hits the brake pedal. It takes another 8.5 seconds for the car to stop. (a) How far did the car travel from the time you first saw the red light until it stopped? (b) What was the average velocity of your car during this period? (c) What was the average acceleration during this period?

a) During the first 0.5 s the velocity is constant, so the car travels 0.5s x 100km/hr = 0.5s x 100,000m/3,600s = 13.9m. During the next 8.5 seconds the car is accelerating with a = (0km/hr - 100km/hr)/8.5s = (-100,000m/3,600s)/8.5s = -3.27m/s². Thus, the car travels an additional 100,000m/3,600s x 8.5s -(3.27m/s²)/2 x (8.5s)² = 118m. So the total distance traveled is 131.9m.

b) v_ave = 131.9m/9s = 14.7m/s = 52.8km/hr.
c) a_ave = (0km/hr - 100km/hr)/9.0s = (-100,000m/3,600s)/9.0s = -3.09m/s²

Problem #2)
You want to row across a small river that is flowing at 4 km/hr in the shortest possible time. You can row at 6 km/hr. (a) What direction should you row (relative to the bank of the river)? (b) If the river is 120 m wide, how long will it take you to cross? (c) Relative to your starting point, where do you reach land on the opposite bank?

a) To cross the river in the shortest possible time, you need to maximize your velocity component in the direction perpendicular to the riverbank. You can do this by rowing in a direction perpendicular to the bank (90 deg). (Note that because of the current your direction of travel will not be directly across the river.)

b) Since your velocity component perpendicular to the river is 6km/hr, the time it will take you to cross the river is 120m/(6km/hr) = 120m/(6,000m/3,600s) = 72s = 1.2min.

c) During this time you are being carried downstream by the current. So you land a distance (4km/hr)(1.2/60)hr = 0.08km = 80m downstream from your starting point.

Problem #3)

The ball in the diagram on the left is moving horizontally with a velocity v on the taller block. The trajectory of the ball after it leaves the taller block is such that the very bottom of the ball lands on the near edge of the shorter block. (a) How fast was the ball moving when it left the taller block? (b) How fast was it moving when it hit the shorter block?

a) Since v_0y = 0, the time that it takes the ball to drop 2m can be found from -2m = -(g/2)t², which yields t = 0.639s. For the ball to land on the edge of the smaller block, it also must travel a distance of 2m horizontally in this same time. I.e., v_0x(0.639s) = 2.0m, which implies that v_0x = 3.13m/s.

b) To find the speed at which the ball hits the smaller block, we need to know both its horizontal and vertical velocity components. The horizontal component is just v_0x, the vertical component can be found from v_y = -gt = -9.8m/s²(0.639s) = -6.26m/s. So the speed is given by [(3.13)² + (-6.26)²]^1/2 = 7.00m/s.

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