Physics Exercises/Kinematics in 2-3 Dimensions/Solutions

Without Calculus

1. The most important thing to notice in this problem is that there is no force in the horizontal direction: so there is no horizontal acceleration, and therefore the horizontal velocity is constant.
   1. From trigonometry (draw a figure and find the right triangle with) $v_{x}=v_{0}cos(\theta )$.
   2. Same as above, $v_{y}=v_{0}sin(\theta )$.
   3. Because $a_{x}=0$, $v_{x}=v_{0}cos(\theta )$.
   4. Note that at the top, the cannon ball is neither moving higher or lower (by definition), so the vertical velocity must be: $v_{y}=0$.
   5. Again, because $a_{x}=0$, $v_{x}=v_{0}cos(\theta )$.
   6. One easy way to find this is by symmetry. In general (esp. if the final height is not the same as the initial height), we can use conservation of energy (but you most likely haven't learned this yet). $v_{y}=-v_{0}sin(\theta )$. The negative sign denotes the downward direction.
2. This problem uses what is called Galilean relativity. Arithmatics used here are approximately correct only at speeds much less than speed of light, c, which is correct for every day objects.

   1. 5 km/h. Keep in mind that velocity is a vector, and add two velocities vectorially. By the Pythagorean theorem, her speed relative to shore is given by:

      $v={\sqrt {v_{\mathrm {boat} }^{2}+v_{\mathrm {walk} }^{2}}}$.

      Plugging in the given numbers yields the answer above.

   2. 53.1 deg. The easiest way to see this is by drawing two vectors tails together (so that you can see the triangle). The angle, which is the angle between the sum of two vectors and the direction north, can be found in terms of inverse tangent of ratio of the legs: $\theta =\arctan(4/3)$, which has the approximate value given above.

With Calculus

1. This problem is more or less a calculus exercise (which is why we told you to ignore units for this problem—it's not very physical).

   1. Remember speed is the scalar quantity equal to the magnitude of the velocity vector. Use the Pythagorean theorem to calculate the magnitude:

      $v(t)={\sqrt {(sin(t))^{2}+e^{2t}}}$.

   2. For this, remember that in general,

      $x(t)=\int _{0}^{t}v(t)dt'$.

      Performing the integral separately for each perpendicular direction gives:

      ${\vec {x}}(t)=-cos(t)~{\hat {x}}+e^{t}~{\hat {y}}$.

      Here, the integration constant is zero because we start out at origin.

   3. Remember that acceleration is also a vector quantity. This is same as above, except this time, you perform a derivative, since ${\vec {a}}={\frac {d{\vec {v}}}{dt}}$. After the differentiation, we get:

      ${\vec {a}}(t)=cos(t)~{\hat {x}}+e^{t}~{\hat {y}}$.