Assignment:
1) An American football player, starting from rest at the line of scrimmage, accelerates along a straight line for a time of 1.25 s. Then, during a negligible amount of time, he changes the magnitude of his acceleration to a value of 1.2 m s-2. With this acceleration, he continues in the same direction for another 1.2 s, until he reaches a speed of 3.7 m s-1. What is the value of his acceleration (assumed to be constant) during the initial 1.25 s period?
i. Sketch a plot of velocity versus time.
ii. To solve this problem, you have to work backwards. If the final speed is 3.7 m s-1after accelerating at 1.2 m s-2 for 1.2 s, then what velocity did he have before this 1.2 s burst of acceleration?
iii. It took the player 1.25 s to reach this velocity, accelerating from rest. What was the initial acceleration?
2) Two students are canoeing on a river. While heading upstream, they accidentally drop an empty bottle overboard. They then continue paddling for 1 hour, reaching a point 2 km farther upstream. At this point they realize that the bottle is missing and, driven by ecological awareness, they turn around and head downstream. They catch up with and retrieve the bottle (which has been moving along with the current) 5 km downstream from the turn-around point. Assuming a constant paddling effort throughout, how fast is the river flowing? What would the canoe speed in a still lake be for the same paddling effort?
The unknown variables in this problem seem to be the speed of the river, the speed of the canoe relative to the river and the time it takes from when they turn around to when they retrieve the bottle. 3 unknowns.
i. Write down an equation in terms of the 3 unknowns for the students canoeing upstream. This is of the form, velocity is equal to distance over time.
ii. Write down a similar equation in terms of the 3 unknowns for the students canoeing down stream to retrieve the bottle.
iii. Write down a similar equation in terms of the 3 unknowns for the journey
iv. You now have 3 equations and 3 unknowns. Do the algebra and solve the problem.
3) The mass of an ant is 5.5x10-6 kg. What is this in grams (g), milligrams (mg) and micrograms ( g)? A dose of a given antibiotic for an infant is one micro-liter per hour (1 l h-1). What is this in meters cubed per second (m3 s-1) and liters per year (l year-1)?
To change units simply multiply by a conversion factor. For example, to convert 2.6 cm to m would be
2.6cm (1m /100cm) = 0.026m
where the fraction is the conversion. Note 100 cm is equivalent to 1 m, so multiplying the 2.6 cm by this fraction doesn't change the quantity just the units.
4) You are driving along with a furry dice hanging from the ceiling of your car. You observe that the furry dice are motionless relative to the car. Draw a clearly labeled free-body diagram for the furry dice if your car has a uniform velocity. Draw a clearly labeled free-body diagram for the furry dice if your car is speeding up uniformly.
Recall that a free-body diagram (FBD) is a diagram where the body (here furry dice) is represented by a dot and the forces acting on that body are represented by arrows emanating from that dot. What forces are acting on the furry dice?
By "motionless relative to the car" the question is telling you that the furry dice are not swinging around, by "uniform velocity" the question tells you there is no acceleration here and the net force is equal to zero (in both x- and y- directions). Finally, by stating that the "car is speeding up" it now implies that the dice, which recall are "motionless relative to the car", must also be accelerating and for this to happen there must be a net force in the direction of acceleration.
5) A firefighter who weighs 712 N slides down a vertical pole with an acceleration of 3 m s-2, directed downward. What are the magnitude and direction (up or down) of the vertical force on the firefighter from the pole and the magnitude and direction of the vertical force on the pole from the firefighter?
i. Again this is a problem which requires a FBD. Draw a FBD for the firefighter. What forces are acting on him, and what must the net force on the firefighter be for his/her acceleration to be 3 m s-2
ii. Once you've found the force on the firefighter from the pole, the force on the pole from the firefighter is easy, right? Look up Newtons' 3rd law.
6) The coefficient for static friction for rubber on dry asphalt is from 0.35 to 1.2 (average of say 0.775), while for rubber on wet asphalt its from 0.25 to 0.8 (average of say 0.525). These values are taken from Baker, J.S., "Traffic Accident Investigation Manual", 1975.
Consider a car traveling at 20.1168 m s-1 (45 mph) with a driver reaction time of 0.75 s on a dry road.
At what speed should the driver travel in wet conditions to maintain the same stopping distance? Why is the coefficient of static friction used here? Why not the coefficient of kinetic friction?
i. First find the stopping distance for the car on a dry road. Note that there are two parts to this. First the car moves at a constant velocity for 0.75 s (because the car doesn't decelerate until after the reaction time) and then it decelerates linearly from the initial velocity to the final velocity (zero because it stops). There are different ways to solve this. You could write down an equation for distance with two terms. The first term would be distance during the reaction time is velocity multiplied by time.
The second term would have to take into consideration the deceleration (and the coefficient for static friction). Alternatively, you could sketch a plot of velocity versus time (constant for 0.75 s and then decreasing linearly to zero with a slope equal to deceleration). How do you get distance from a velocity vs time graph?
ii. Repeat what you just did, but in reverse. Now you know the stopping distance (from the first part) and you do the same math (or use the same kind of velocity vs time graph) to find the initial velocity.
iii. Why static and not kinetic friction? Google ABS breaks.