QUIZ 2 (ANSWERS)
READ ALL EXAMPLES. SOME PROBLEMS ARE JUST LIKE AN EXAMPLE---SEE EXAMPLE 2.9 AND #66. 
CH. 2 acceleration problems without Free Fall ( Part I)
CH. 2 - Multiple choice 1,2,3,7,12,13, and exercise/problems 1*,2*,8*,15*,20*,67*,25*,26*,37*,38*,40*,76*,80* ,66*,68*
* DISCUSSIONS PROVIDED. 
SELECTED DISCUSSIONS TO exercises/problems BELOW.

ALL PROBLEMS ARE DUE EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED PROBLEMS.
1. I WOULD EXPECT YOU TO BE ABLE TO DO THIS PROBLEM WITHOUT MUCH EFFORT: 
HERE'S A HINT:
(a) THE ANT'S POSITION AT A IS NEGATIVE, AND AT C AND B, POSITIVE.
(b) The displacement may be negative or positive. For example in the case of B to C, displacement is -30 cm.
2. (a) Displacement , as opposed to distance discussed in detail  #80 below, is defined to be xf - xi.   Here f means final and i means initial.  Compute the difference for each requested time interval.  For example, between 1.0 s and 10.0 s, the displacement is 6 m -  1 m.
(b) FOR DISTANCE you must find the ABSOLUTE VALUE of each displacement. Note the displacement can be positive or negative, but distance is always positive,  so you must find the absolute value
Below is another way of seeing the problem:

Solve:  (a) The displacement is (i) (ii) (iii)  (iv)

(b) (i)  (ii)  (iii) zero (stays at )


8. In each case recall the slope of x is the velocity   vx.  (a) The slope  constant and greater than zero. (b) slope is zero (c)  slope is negative and less than zero. 
Below is another way of seeing the problem:

 is the slope of the x versus t graph. In each case this slope is constant, so  is constant.

Solve:  The graphs of  versus t are sketched in the figure below.


15. (a) See example 2.12. The relative speed is given by the ratio of the truck length and the time to pass.  Since you are going faster than the truck and in the same direction,  subtract the relative speed from your speed to get  truck's speed relative to road.  
(b) distance = speed * time, where you are using speed relative to the road and time to pass. 
20. (A) the slope of constant; find rise/ run (B) same slope as previous part. (C) the slope is zero (D) slope is constant and negative; find rise/run  (E) same slope as previous part (F)  same slope as previous part (G) the slope of the tangent is zero 
 Below is another way of seeing the problem:


2.20.  Set Up:  The instantaneous velocity at any point is the slope of the x versus t graph at that point. Estimate the slope from the graph.

Solve:  A: B: C: D: E: F: G:

Reflect:  The sign of  shows the direction the car is moving.  is constant when x versus t is a straight line.

67.  Average speed = total distance /total time, where total time = 10 mi/8 mi/h +  10 mi/V and V is the average speed on the last half of the trip.  We have total average speed = 20 mi/( 10/8 h  +  10/V) .  For each part set total average speed to the listed value and solve the previous equation for V. 
(a)   total average speed = 4 mi/h.
(b)  total average speed = 12 mi/h.
(c)  total average speed = 16 mi/h.
V must be a positive number; otherwise there is no solution.
25.  In all cases remember the acceleration is the slope of the velocity.  (a) slope is positive and constant (b) slope is positive and  increasing (c) slope is negative and constant. (d) slope is positive and decreasing with time. 
26. (a) read from the graph (b) acceleration = slope of velocity (c) see 25 (c).

Below is another way of seeing the problem:

The acceleration  is the slope of the graph of  versus t.

Solve:  (a) Reading from the graph, at to the right and at to the left.

(b)  versus t is a straight line with slope  The acceleration is constant and equal to  to the left.

(c) The graph of  versus t is given in the figure below.

 

 

 

 

37. See discussions for # 76. For part (c) compute the distance  20 m/s traffic moves for the time period easily computed in part (b).  To do this problem select from the set of equations 2.8 to 2.14.
38. Below is another way of seeing the problem:

 The car travels at constant speed during the reaction time. Let  be the direction the car is traveling, so  after the brakes are applied.

Solve:  (a) During the reaction time the car travels a distance of  

For the motion after the brakes are applied,    and    gives

The total distance is

(b)  A calculation similar to that of part (a) gives a total stopping distance of

40.  Below is another way of seeing the problem:

Set Up:  Let  be the direction the train is traveling. Find  for each segment of the motion.

Solve:    to 14.0 s:  At  the speed is

In the next 70.0 s,  and  For the interval during which the train is slowing down,    and    gives

The total distance traveled is


66.  SEE EXAMPLE 2.9 ! The equation for the truck is x = (15 m/s)*t  and for the car x = (1/2)*(2.5 m/s2)*t2.  When the car overtakes the truck the two values of x are equal, so set the expressions equal  to each other and find t, then find x. You can find the velocity of the car when it it overtakes the truck by using equation 2.8.
Below is another way of seeing the problem:


Set Up:  At  the auto and truck are at the same position. The auto overtakes the truck when after time T they have both traveled a distance d.

Solve:  (a) Apply  to the motion of each vehicle. The auto has  and  so  The truck has  and  so  Combining these two equations gives  and  Then

(b)

76. You  want the ramp to be long enough for less powerful cars and for cars with bald tires. These are you standards from which road specs are found.
#Find the acceleration of the less powerful car using equation 2.8 AND 60 MPH.  Then use equation 2.13 to find the ramp length by setting the final velocity equal to 70 mph assuming the car starts from rest.  
# Find the acceleration of the  car with bald tires using equation 2.8 AND 60 MPH. Your acceleration will be negative assuming the car moves in the positive direction when braking  Then use equation 2.13 to find the ramp length assuming the car starts at 70 mph and finally comes to rest..  
Below is another way of seeing the problem:

Set Up:  Take  to be the direction the car moves when speeding up. Use the acceleration and stopping data to find a in each case.  Design the on-ramp for the less powerful car and the off-ramp for the car with bald tires.

Solve:  on-ramp (speeding up):      and  gives

 

Then    and  in  gives

This is the required length of the on-ramp.

off-ramp (slowing down):      and  gives

 

Then    and  in  gives

 

This is the required length of the off-ramp.

68. Find the time it takes to run 10,000 m at 2.5 m/s, then subtract 15 minutes from your answer.  With  the shorter time period you computed, finally find the speed of yourself if you run 10,000 m.
Below is another way of seeing the problem:

. Set Up:  Both you and your friend run at a steady pace over the 10 km race so the equation  applies. You can calculate the time that it takes for your friend to complete the race and then subtract 15 minutes (900 seconds) to obtain the time that you need to achieve your goal. The race is 10 km, which is

Solve:  The time your friend takes to complete the race is  Thus, you need to complete the race in  The required speed to do this is

 

80.
(a) The mouse is to the right of the origin wherever the value of x is greater than zero.
(b) Find the slope of the first straight line segment  defined between 0 and 3 seconds.
(c) Wherever the shape of the x- graph is a straight line, the acceleration  is zero since the velocity is constant on those line segments .  Does the x- graph ever become curved instead of straight? If so,  the acceleration has become non zero. The x-graph seems to be a mixture of straight line segments and curves so the acceleration seems to change during the entire trip
(d) The maximum speed is given by the section of the graph with the largest magnitude of slope---either the slope  of a straight line segment OR the  slope of the tangent to the graph.    Where does this occur?
(e) If the mouse moves rightward, the slope is positive. Otherwise it moves either leftward (if there is a negative slope) or does not move (wherever the slope is zero.)
(f) Observe the graph. To get the distance you must always add since distance is always positive. In the first 3 seconds for example, the mouse moves 40 cm. For the entire 10 seconds, you must find the individual  distances for EACH  time period AND ADD THEM UP. Hints: Here are helpful  time periods: 0 to 3, 3 to 5, 5 to 7 and 7 to 10 seconds.  LOOK AT EACH TIME INTERVAL AND FIND THE DISTANCE FOR EACH  FROM THE GRAPH, THEN ADD THEM UP.
(g) The mouse speeds up whenever the slope seems to get steeper.  Look at the curved sections of the graph or wherever the slope changes.
(h) Those equations  apply wherever the curve is a straight line segment or is parabolic.
Below is another way of seeing the problem:

:  The velocity is the slope of the x versus t graph. The sign of the slope specifies the direction of the velocity.

Solve:  (a) The mouse is to the right of the origin when x is positive. This is for  and for  The mouse is to the left of the origin when x is negative. This is for  The mouse is at the origin when  6.0 s, 8.5 s.

(b) At  is positive and has magnitude

(c) The acceleration is not constant. For constant a the x-t graph would be a section of a parabola, and it is not.

(d) The speed is greatest when the magnitude of the slope is greatest, and this is between 5.0 s and 6.0 s. The speed in this interval is

 

During this part of the motion the velocity is negative and the mouse is moving to the left.

(e) The mouse is moving to the right when x is increasing. This happens for  and for  The mouse is moving to the left when x is decreasing. This happens for  The mouse is instantaneously at rest when the slope of the tangent to the x versus t graph is zero, when the tangent is horizontal. This happens for  and at

(f) In the first 3 seconds the mouse travels from 0 to 40.0 cm so it travels 40.0 cm. In the first 10 seconds the mouse travels from 0 cm to 40.0 cm, from 40.0 cm to  and from  to 15.0 cm. The total distance traveled is

(g) The mouse is speeding up when the slope of x versus t is increasing. This happens at 5.0 s and between 7.0 s and 8.5 s. The mouse is slowing down at 3.0 s, between 6.0 s and 7.0 s and between 8.5 s and 10.0 s.

(h) No, the acceleration is not constant during this entire interval and these formulas apply only for constant acceleration.