Kinematics 1 Problem Solutions

Each solution provided below is not necessarily the only possible solution.  If a solution differs greatly from yours, copy and paste it from your browser to your Maple worksheet, then execute it and see if the final answers agree.

Quick access to the solutions of the problems (2a), (2b), (3), (4), and (5).

(1a)  7875 m/s

restart;
eq := d = r*t;
d := 525*(.5);
t := 1/30;
eq;
solve( eq, r );      This yields the speed.  The last two lines are a check.
subs( r = 7875, eq );
evalf( 7875*t );

(1b)  21.2 m/s

The definition of average velocity:

In this case the numerator is the hypotenuse of a right triangle with sides of .5 m.  Since we only want the magnitude of the average velocity and the change in time is positive, we may use the absolute value of the numerator, which is just the length of the hypotenuse.  Thus,

vel[avg] := sqrt( .5^2 + .5^2 )/ (1/30);                               Check out ?sqrt.

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(2a)  .1 yrs

restart;
vel := v[f] = v[i] + a*t;
v[f] := (1/10)*(3*10^8);
a := 10;
v[i] := 0;
vel;
t := solve( vel, t );
vel;      The above assignment of t makes this statement a check.
years := t*(1/60)*(1/60)*(1/24)*(1/365);
evalf(years);

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(2b)  4.5 x 10^13 m

distance := x[f]-x[i] = ( (v[f]+v[i])/2 )*t;
subs( x[i] = 0, distance );
t := 't';
pos = (1/2)*a*t^2;      The live plot of this equation is not very pretty.   The following sections of the tutorial show some techniques for improving its appearance.  Of course, you may rescale the horizontal and vertical axes by right clicking on the plot and resetting the appropriate parameters.

(3)  3.2 m

restart;
pos := y[f] = y[i] + v[i]*t + (1/2)*a*t^2;
y[i] := 50;
y[f] := 0;
v[i] := 0;
a := -10;
solve( pos, t);
t := evalf( sqrt(10) );
pos;
pos2 := d = r*t;
r := 1;
pos2;

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(4)  .23 m

Convince yourself that the solution is the same if the sound wave starts 2 m away from the other side of the wall and moves towards the bat.  The following solution is modeled on this assumption, placing the initial position of the bat 2 m to the left of the wall and the initial position of the wave 2 m to the right of the wall.

restart;
bat[f] := -2 + 20*t;
wave[f] := 2 + (-330)*t;
solve( bat[f] = wave[f], t );
t := evalf( 2/175 );
bat[f] = wave[f];
distance[bat] := bat[f] - (-2);

(5)  The boy gets trampled.

The following solution finds the times at which the boy and the bulls reach the same position.  It sets the position equations of the boy and the bulls equal to each other and then finds the times at which the resulting quadratic equation has solutions, i.e. at these times the position of the bulls and the position of the boy will be the same.  If there is a real solution, then the boy is trampled.  If there is no real solution, then the boy survives.

restart;
a[boy] := 3;
pos[boy] := 20 + 0*t + (1/2)*a[boy]*t^2;
pos[bulls] := 0 + 20*t + (1/2)*0*t^2;
solve( pos[boy] = pos[bulls], t );

Note that you may select the two single variable expressions below and place them into the same set of axes, revealing a graphical solution to the problem.  The ranges on the axes in the plot below were obtained by right clicking in the plot and resetting the vertical range to 0-80, and leaving the horizontal range on default.

> pos[boy];

> pos[bulls];

Kinematics 1 Problem Set

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