Kinematics In Two Dimensions Practice


Kinematics In Two Dimensions 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…

  1. the total distance of the entire trip
  2. the total displacement of the entire trip
  3. the average speed of the entire trip
  4. the average velocity of the entire trip
  5. the average acceleration of the entire trip


  1. 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

  2. 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

  3. 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
    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

    Δt1 =
    Δt1 =
    6.0 km
    6.0 km/h
    Δt1 =
    1.0 h
    Δt2 =
    Δt2 =
    10 km
    5.0 km/h
    Δt2 =
    2.0 h
    Δt = Δt1 + Δt2
    Δt = 1.0 h + 2.0 h
    Δt = 3.0 h

    The average speed is then…

     =16 km
    3.0 h
     = 5.3 km/h
  4. 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.
     =11.6619… km
    3.0 h
     =3.8873… km/h
     = 3.9 km/h 59° W of N
  5. 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…

  1. the speed of the current
  2. the magnitude of the swimmer’s resultant velocity
  3. the direction of the swimmer’s resultant velocity
  4. the time it takes the swimmer to cross the river


Since distance and velocity are directly proportional, this begins as a similar triangles problem.

  1. 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
    40 m =vcurrent
    80 m1.6 m/s
    vcurrent = 0.8 m/s
  2. 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]
    v = 1.8 m/s
  3. 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
    tan θ =40 m =0.8 m/s
    80 m1.6 m/s
    tan θ =0.5
    θ = 27° downstream
  4. 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
    t =40 m =80 m
    0.8 m/s1.6 m/s
    t = 50 s

    It also works along the resultant direction…

    t =r
    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 one-dimensional equations. This idea is central to the field of analytical geometry.

practice problem 3

A car enters an intersection at 20 m/s where it collides with a truck. The impact rotates the car 90° and gives it a speed of 15 m/s. Determine the average acceleration of the car if it was in contact with the truck for 1.25 s.


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
 =Δv =25 m/s = 20 m/s2
Δt1.25 s
tan θ =15 m/s
20 m/s
θ =143°

 = 20 m/s2 at 143°

practice problem 4

The asteroid 2007 VK184 is classified as a near earth object (NEO). It has an orbit that brings it close enough to earth, often enough that we need to be concerned. It will make 7 close approaches in the Twenty-first Century. The first occured in 2007 when the asteroid was discovered (thus the provisional name 2007 VK184). The last will be in 2048 when there is a small but non-zero probability (< 1%) of collision with the Earth. This is the only asteroid currently known to pose a threat to the Earth in this century. On 30 May 2048 at 22:11 UTC (6:11 PM New York time), asteroid 2007 VK184 will be traveling at an estimated 16.9401 km/s relative to the sun. If it strikes the Earth, the speed of impact would be 16.9436 km/s (Source: JPL). Determine the impact angle between the velocity vectors of the Earth and the asteroid. Kinematics


Start with a diagram.

animated collision

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…

v =s
v =2π(1.50 × 1011 m)
(365.25 × 24 × 60 × 60 s)
v =29.8653 km/s

or use an online calculator (which knows the average earth-sun distance with more precision)… Kinematics

v =s
v =2π(1 astronomical unit)
(365.25 day)
v =29.7853 km/s

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
− 2bc
cos A =(16.9 km/s)2 − (29.8 km/s)2  − (16.9 km/s)2
− 2 (29.8 km/s) (16.9 km/s)
A = 28.5°

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