Study Guide 9 - Kinematics

What is the relation between constant speed (v), displacement (d) and time (t)?

\(\mathrm {v = d/t = 10\times 10^3\; m/(15 \times 60) s = 11\; m/s}\)

The displacement is defined as: final position - initial position

So in this case the displacement is zero.

Since from part a the displacement is zero then the average velocity is also zero.

Let \(\mathrm {t}\) be the elapsed time from when the first rider starts until the second rider catches up.
 
\(\mathrm {Therefore\; x_1 = 10t}\)
 
For the second rider who starts 10 s later
 
\(\mathrm {x_2 = (t - 10) 12}\)
 
but when they meet, \(\mathrm {x_1 = x_2}\)
 
\(\mathrm {(t - 10) 12 = 10 t}\)

\(\mathrm {t = 60 \;s}\) or \(\mathrm {50\;s}\) after the second rider starts.

You must use one of the Gallilean equations. In this case \(\mathrm {v = v_0 + at}\)

\(\mathrm {v = v_0 + at \\ 10 = 5 + a(5) \\ a = (10 - 5)/5 = 1 m/s^2}\)

In the first 2 seconds his velocity is: Constant

So the acceleration is zero since the slope of the s vs. t curve is constant.
\(\mathrm {v = \Delta s/\Delta t = 4/2 = 2 m/s^2}\)

In the period 2 to 4 seconds his position is: Constant

Therefore his velocity is zero.

Is the velocity changing in the first 2 seconds? No

The velocity is not changing; it is constant at 2 m/s, so a = 0

Is the velocity changing in the period 2 to 4 seconds? No

The velocity is not changing; it is constant at a value of \(\mathrm {0\; m/s,}\) so \(\mathrm {a = 0}\)

Is the velocity changing in the period 4 to 6 seconds? Yes

The velocity is the slope of the s vs. t curve, i.e., \(\mathrm {v = ds/dt.}\) This slope is clearly changing in the period 4 to 6 seconds, so the acceleration is non-zero.

\(\mathrm {x_m = v_mt = 25t \\ x_p = v_{0_p}t + ½(a)t^2 = 10t + ½(10)t^2 \\ but\; x_m = x_p \\ 25t = 10t + ½(10)t^2 \\ t = 3 \;s}\)

\(\mathrm {x = v_0t + ½at^2 \\   75 = 15t + (9.8/2)t^2 \\ 4.9t^2 + 15t - 75 = 0 \\   t = -(15/9.8) ± (1/9.8)[15^2 + 4(4.9)(75)]^{½} \\   = (1/9.8)(-15 ± 41.2) Choose\; t\; positive \\   t = 26.2/9.8 = 2.67\; s}\)

\(\mathrm {v^2 = v_{0}\;^2 + 2ax = 4^2 + 2(9.8)2 = 55.2 \\ v = 7.4 \;m/s}\)

a. is the correct answer


If you have difficulty with the addition of vectors you are strongly urged to use the Vector tutorial.

c. is the answer

In the standard x, y coordinate system the x component over the hypotinuse is given by the cos of 53.1

\(\mathrm {So\; F_x = F \cos \;53.1^\circ = 20 \times 0.60 = 12 \;N.}\)

It follows that the y component is given by the sin of 53.1.

\(\mathrm {F_y = F \sin\; 53.1^\circ = 20 \times 0.80 = 16 \;N}\)

What forces act on the ball in the x-direction after it is released?
There is no force on the ball in the x-direction. Therefore the motion of the ball in the x-direction is motion at constant velocity \(\mathrm {(a_x = 0).}\)

What forces acts on the ball in the y-direction?
The force of gravity acts in the y-direction so the motion in the y-direction is motion at constant acceleration
with \(\mathrm {a_y = -g.}\)

What is the velocity in the x-direction?
\(\mathrm {The\; velocity\; in\; the\; x-direction\; is\; +v \cos \;45^\circ}\)

What is the initial velocity in the y-direction?
\(\mathrm {The\; initial \;velocity \;of \;the \;ball\; in \;the\; y-direction\; is\; v \sin \; 45^\circ}\)

We can now proceed to solve the problem:

What is the displacement in the y-direction at the end of time t?
(a) 100 m
(b) 2 X the height of the path
(c) 0

\(\mathrm{v = v_0 + at \\ 20 = 10 + a(10) \\ a = 10/10 = 1\; m/s^2}\)
 
\(\mathrm {x = v_0t + ½at^2 = 10(10) + ½(1)10^2 = 150 \;m}\)
 
\(\mathrm {s_x = 40 + 150 \sin\;30 = 40 + 75 = 115 \;m}\)
 
\(\mathrm {s_y = 150 \; \cos \;30 = 130 \;m}\)
 
\(\mathrm {s = (115^2 + 1302)½ = 173 \;m}\)
 
\(\mathrm {\theta = \tan^{-1} (130/115) = 48.5\; N \;of \;E}\)

\(\mathrm {\theta = \sin^{-1}(3/10) = \sin^{-1}0.3 = 17.5^\circ}\)

The motorboat must travel across the river with speed v and also upstream at 3 m/s to counter the flow of the river. Since the boat's speed is 10 m/s then:

\(\mathrm{ 10^2 = v^2 + 3^2 \\ V^2 = 100 - 9 \\ V = 9.54\; m/s}\)

\(\mathrm {v = (20^2 + 15^2)^{1/2} = 25\; km/hr}\)
\(\mathrm{\theta = \tan^{-1} (15/20) = 37 \;degrees\; E\; of \;N}\)

The airplane must fly 30º E of N at a speed of 500/2 = 250 km/hr.

From the drawing we can see that the resultant velocity of the plane, \(\mathrm {v,}\) will be the vector sum of the velocity of the plane in still air, \(\mathrm {v_p,}\) plus the velocity of the air itself, \(\mathrm {v_w,}\) (i.e., the wind speed).

Therefore: \(\mathrm {v = v_p + v_w}\)

But this means that the x and y components also add individually. In addition, the magnitude of the v vector must be 250 km/hr (see part (a)) and the magnitude of the \(\mathrm {v_w}\) vector is 60 km/hr.

First evaluate the x components:
\(\mathrm {v_x = v_{px} + v_{wx} \\ 250\; \cos \;60 = v_{px} + 60\; \cos\; 30 \\ v_{px} = 73\; km/hr}\)

Now evaluate the y components:
\(\mathrm {v_y = v_{py} + v_{wy} \\ 250\; \sin \;60 = v_{py} + 60\; \sin \;30 \\ v_{py} = 187\; km/hr}\)

\(\mathrm {V_p = ({v_{px}}^2 + {v_{py}}^2)½ = 200\; km/hr}\)

\(\mathrm {\theta = \tan^{-1}(73/187) = \tan^{-1}0.39 = 21.4^\circ \; E\; of\; N.}\)