Physics 12 Practice Quiz for Kinematics Solutions

 

Some notes to explain the equations:

 

• The symbol D means “change in” so that DT is the change in T (whatever T represents).  The change in something is the difference between what it is at the end, or finally, T_f and what it was at the start, or initially, T_i  DT = T_f - T_i

 

• I can’t do vector arrows on the computer so I will use another notation that is common.  Bold symbols represent vectors, so p is a vector quantity while p is a scalar quantity.

 

• s  represents the position of an object and t represents a time.

 

Equations:

 

Definitions

1. Elapsed time = Dt = t_f - t_i
2. Displacement = Ds= s_f - s_i
3. Average velocity = v = Ds /Dt
4. Average acceleration =  a = Dv /Dt  (where, Dv = v_f - v_i)

 

For an object with constant acceleration = a

1.     s = v_o t – ½at²

2.     v = v_o  – at

 

For an object traveling with only gravity acting on it (down is negative, g = 9.8m/s²)

 

	Acceleration	Velocity	Position/Displacement
Horizontal	ax = 0 m/s2	vx = vocosθ	x = vocosθ t
Vertical	ay = – g = – 9.8 m/s2	vy = vosinθ –  g t	y = vosinθ t – ½g t2

 

1. If the object is launched with no initial velocity or only upward or downward velocity, then we pick θ=90º
2. If it has initial velocity all in the horizontal direction (as in rolling off a table, a rifle shooitng horizontally, etc.) we set θ=90º

 

 

1)      A runner runs from Laguna Alalay to the Mormon Temple which is 3.5 km to the north in 10 minutes.

a)      What is their displacement?

Ds =3.5km North

 

b)      What is their average velocity?

v = Ds /Dt

= 3.5km North/600s

= 3500mN/600s

= 5.8m/s North

 

c)      If they are traveling at 2.0m/s  3.0 seconds before they arrive at the Temple and come to a stop at the end of their run, what is their average acceleration for the last 3.0 seconds?

a = Dv /Dt 

= -2.0m/s / 3.0s (calling North positive))

= -0.67m/s²

=  0.67m/s² South

 

2)      A ball is thrown with an purely upward velocity of 2.5 m/s from a height of 50.0 m:

a)      How high will the ball go?

This is physics code for “at what height will the vertical speed go to zero.”

 

The y velocity equation will be (see above table):

 

v_y = 2.5m/s – gt = 0 when

t = 2.5m/s /g = 2.5/9.8 = 0.255 s

 

The y equation gives height, y, as a function of time

 

y = 2.5m/s × t - ½gt² and we use t from above to get

y = 2.5×0.255  – 4.9× (0.255) ²

= 0.319 m

 

This is the height it goes ABOVE the 50.0m it started at so it gets to 50.3m

 

 

b)      How long will it take to hit the ground?

 

There are a number of ways to do this but the easiest way is to ask at what time the ball will have a position of -50.0m (down is negative and the ball starts at zero)

The equation for y is given above as

y = 2.5m/s × t - ½gt²

and when y = -50m we get

-50 = 2.5t – 4.9t²                                 (dropping units and using g=9.8)

            rearrange to get

                        4.9t²-2.5t -50 = 0                     (correct quadratic form)

            Solve it for t:

                        t = (2.5 ± sqrt(6.25+980))/9.8 =  3.5s and -2.9 s  

(we want positive time in the future)

 

c)      With what velocity will it hit the ground?

The velocity equation is:

v = 2.5 – gt and we can use the time from b)

v = 2.5 – 9.8×3.5 = -31.8 m/s, or 31.8m/s DOWN

 

3)      A ball is thrown with an initial velocity of 20 m/s at an angle of 25º above the horizontal:

a)      How long until it reaches a height of 1.0 m?

Using the table equation y = v_osinθ t – ½g t²

with v_o= 20m/s and θ = 25º  and set y = 1.0m

Solve for t:

1.0 = 20sin25º t – ½g t²

 

Rearrange to make it look like a proper quadratic and solve it for t.  You get two roots to the quadratic, t = 0.12s and 1.5s.  This represents the ball going on the upward part of the parabolic arc first and then coming down through a height of 1.0 m again.  TAKE t=0.12s AS YOUR ANSWER

 

 

b)      How far will it have traveled horizontally at this point?

Use the x equation, x = v_ocosθ t

with the time from above, t= 0.12s and  with v_o= 20m/s and θ = 25º 

 

                        You get x=2.1m, I think

 

c)      What will its velocity be at this point?

Use the velocity equations in the x and y direction

v_x = v_ocosθ   and v_y = v_osinθ –  g t

 with the time from above, t= 0.12s and  with v_o= 20m/s and θ = 25º 

 

This gives the components of velocity and then you must use Pythagoras and SOHCAHTOA to get the velocity in magnitude/direction form.  Why don’t one of you do the calculations and e-mail them out to the class list…