practice problem 1
- the total distance of the entire trip
- the total displacement of the entire trip
- the average speed of the entire trip
- the average velocity of the entire trip
- the average acceleration of the entire trip
- Distance? No problem. First I walked 6.0 km and then I walked 10 km. Thus, I walked 16 km. Distance is a scalar quantity and so the individual distances add just like regular numbers to yield the overall distance…
- Displacement is a bit more challenging, but not by much. Displacement is a vector and vectors have direction, so it's best to diagram this problem (a procedure that's remarkably useful in general). The resultant displacement is the vector sum of the two displacements experienced during the trip. Since they're at right angles to one another, their magnitude can be found using Pythagorean theorem and their direction can be found using tangent…
r = √[(6.0 km)2 + (10 km)2] r = 11.6619… km tan θ = 10 km 6.0 km θ = 59° W of N
- The speed was 6.0 km/h for the first 6.0 km and 5 km/h for the last 10 km. The naive solution is to average the speeds using the add-and-divide method taught in junior high school. This method is wrong, not because the method itself is wrong, but because it doesn't apply to this situation.
6.0 km/h + 5.0 km/h = 5.5 km/h 2 The method of averaging
The weight of the two segments is not equal. The second segment lasted twice as long as the first (as you will soon see).
Go back to the definition to solve this problem. Average speed is the total distance (which we've already found) divided by the total time (which we need to find). Since time is a scalar, all we need to do is find the time for each leg of the journey and add them…
Δt1 = Δs1 v̅1 Δt1 = 6.0 km 6.0 km/h Δt1 = 1.0h Δt2 = Δs2 v̅2 Δt2 = 10 km 5.0 km/h Δt2 = 2.0h Δt = Δt1 + Δt2 Δt = 1.0 h + 2.0 h Δt = 3.0 h
Thus the average speed is…
v̅ = Δs Δt v̅ = 16 km 3.0 h v̅ = 5.3 km/h
- The velocity was 6.0 km/h north over the first 6.0 km and 5 km/h west over the last 10 km. Average velocity is the total displacement divided by the total time. Both of these quantities have already been determined.
v̅ = Δr Δt v̅ = 11.66 km 3.0 h v̅ = 3.8873… km/h v̅ = 3.9 km/h 59° W of N
- Acceleration in this context is relatively meaningless. It would be better to illustrate acceleration in two dimensions with a different problem (like the one below).
practice problem 2
- the speed of the current
- the magnitude of the swimmer's resultant velocity
- the direction of the swimmer's resultant velocity
- the time it takes the swimmer to cross the river
Since distance and velocity are directly proportional, this begins as a similar triangles problem.
- Since speed and distance are directly proportional, the ratio of the downstream distance to the width of the river is the same as the ratio of the current speed to the swimmer's speed.
x = vx y vy 40 m = vcurrent 80 m 1.6 m/s vcurrent = 0.8 m/s
- Determining the resultant velocity is a simple application of Pythagorean theorem.
v2 = vx2 + vy2 v = √[(0.8 m/s)2 + (1.6 m/s)2] 0v = 1.8 m/s
- Direction angles are often best determined using the tangent function. This problem is no exception. The only thing open to discussion is our choice of angle. I suggest using the angle between the resultant velocity and the displacement vector that points directly across the river, but this is just my preference. Be sure to indicate that the resultant lies on a particular side of this vector for clarity.
tan θ = x = vx y vy tan θ = 40 m = 0.8 m/s 80 m 1.6 m/s tan θ = 0.5 θ = 27° downstream
- This is where it gets interesting. By now you should understood that time is the ratio of displacement to velocity. This is a vector problem, so direction matters. This is why we should probably use the words displacement and velocity instead of distance and speed. The only question is which distance and which speed should we use? The simple answer is pick the pair you like the best, just be sure they point in the same direction. It works along either of the component directions…
t = x = y vx vy t = 40 m = 80 m 0.8 m/s 1.6 m/s t = 50 s t = r v t = √[(40 m)2 + (80 m)2] √[(0.8 m/s)2 + (1.6 m/s)2] t = 50 s
practice problem 3
Finding the change in velocity is complicated in this problem by the change in direction. A diagram is indispensable. Let's assume that the initial direction of the car is 0° (to the right in standard position) and that the final velocity will be 90° (toward the top of the page in standard position). The difference of two vectors drawn this way would then connect the the head of the initial vector to the head of the final vector. Use Pythagorean Theorem for magnitude and tangent for direction as usual. Only after we have done all of this can we then plug numbers into the definition.
|Δv =||√((20 m/s)2 + (15 m/s)2) = 25 m/s|
|a̅ =||Δv||=||25 m/s||= 20 m/s2|
|tan θ =||15 m/s|
|θ = 143°|
a̅ = 20 m/s2 at 143°
practice problem 4
Start with a diagram.
Strip it down to its essence.
Two sides of this triangle are given (vasteroid and vimpact). None of the angles are known. The third side (vearth) can be determined from basic knowledge. The average speed of the Earth is the distance covered in one orbit (the circumference) divided by the time it takes to complete that orbit (one year). We could do this on a hand held calculator…
or use an online calculator (which knows the average earth-sun distance with more precision)…
We now have three sides of a triangle and can find the desired angle using the law of cosines.
a2 = b2 + c2 − 2bc cos A
|a =||impact velocity (16.9436 km/s)|
|b =||earth's velocity (29.7853 km/s)|
|c =||asteroid's velocity (16.9401 km/s)|
|A =||impact angle (our goal)|
Solve algebraically, substitute numerical values, and compute the answer.
|cos A =||a2 − b2 − c2|
|cos A =||(16.9 km/s)2 − (29.8 km/s)2
|− 2 (29.8 km/s) (16.9 km/s)|
|A = 28.5°|