discussion  summary  practice  problems 
Kinematics In Two Dimensions Practice
Practice
practice problem 1
I went for a walk one day. I walked north 6.0 km at 6.0 km/h and then west 10 km at 5.0 km/hr. (This problem is deceptively easy, so be careful. Begin each part by reviewing the appropriate physical definition.) Determine…
 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
solution
 Distance? No problem. First I walked 6.0 km and then I walked 10 km for a total of 16 km. Distance is a scalar quantity, so the individual distances add just like regular numbers.
16 km
 Displacement is a bit more challenging. 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 perpendicular to one another, the resultant is the hypotenuse of a right triangle. Its magnitude can be found using pythagorean theorem and its direction can be found using the tangent function.
r = √[(6.0 km)^{2} + (10 km)^{2}] r = 11.6619… km tan θ = 10 km 6.0 km θ = 59° r = 11.7 km
at 59° west of north  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 addanddivide 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 wrong method of averaging The weights of the two segments are 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, add the times for each leg of the journey to get the total time. Kinematics
Δt_{1} = Δs_{1} v̅_{1} Δt_{1} = 6.0 km 6.0 km/h Δt_{1} = 1.0 h Δt_{2} = Δs_{2} v̅_{2} Δt_{2} = 10 km 5.0 km/h Δt_{2} = 2.0 h Δt = Δt_{1} + Δt_{2} Δt = 1.0 h + 2.0 h Δt = 3.0 h The average speed is then…
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 = Δs Δt v̅ = 11.6619… 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
A swimmer heads directly across a river swimming at 1.6 m/s relative to still water. She arrives at a point 40 m downstream from the point directly across the river, which is 80 m wide. Determine…
 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
solution
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 = v_{x} y v_{y} 40 m = v_{current} 80 m 1.6 m/s v_{current} = 0.8 m/s  Determining the resultant velocity is a simple application of Pythagorean theorem.
v^{2} = v_{x}^{2} + v_{y}^{2} v = √[(0.8 m/s)^{2} + (1.6 m/s)^{2}] v = 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 = v_{x} y v_{y} 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 v_{x} v_{y} t = 40 m = 80 m 0.8 m/s 1.6 m/s t = 50 s It also works along the resultant direction…
t = r v t = √[(40 m)^{2} + (80 m)^{2}] √[(0.8 m/s)^{2} + (1.6 m/s)^{2}] t = 50 s There’s an interesting sideline to this question that astute readers might have noticed when looking at the first ratio in the chain of three shown above. The time it takes to cross a river by a swimmer swimming straight across is independent of the speed of the river. The only factors that matter are the speed of the swimmer and the width of the river. This swimmer will always cross the river in 50 s regardless of the speed of the river. 1 m/s, 10 m/s, 100 m/s, it doesn’t matter. This example is a perfect illustration of an idea to be presented in the next section of this book. Motion in two dimensions can be thoroughly described with two independent onedimensional equations. This idea is central to the field of analytical geometry.
practice problem 3
solution
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/s^{2}  
Δt  1.25 s 
tan θ =  15 m/s  
20 m/s  
θ =  143°  
a̅ = 20 m/s^{2} at 143°
practice problem 4
solution
Start with a diagram.
Strip it down to its essence.
Two sides of this triangle are given (v_{asteroid} and v_{impact}). None of the angles are known. The third side (v_{earth}) 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 earthsun distance with more precision)… Kinematics
 
 

We now have three sides of a triangle and can find the desired angle using the law of cosines.
a^{2} = b^{2} + c^{2} − 2bc cos A
Where…
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 =  a^{2} − b^{2} − c^{2} 
− 2bc 
cos A =  (16.9 km/s)^{2} − (29.8 km/s)^{2}  
− 2 (29.8 km/s) (16.9 km/s) 
A = 28.5°  