For a velocity-time graph, the acceleration is determined from a slope calculation. So in each case, two points must be picked and slope calculation must be performed.
a. Pick the two points: (0.0 s, 5.0 m/s) and (5.0 s, 30.0 m/s)
slope = rise/run = (25.0 m/s)/(5.0 s) = +5.0 m/s2
b. Pick the two points: (5.0 s, 25.0 m/s) and (10.0 s, 10.0 m/s)
slope = rise/run = (-15 m/s)/(5.0 s) = -3.0 m/s2
c. From 10.0 to 15.0 s, the slope is a constant value. So determining the slope at 13 seconds is as simple as merely determining the slope of the line choosing any two points. So just pick the two points: (10.0 s, 15.0 m/s) and (15.0 s, 0.0 m/s)
42.
For velocity-time graphs, the displacement of an object is found by computing the area between the line and the time axis. The shape is typically a rectangle (area = base*height), a triangle (area = 0.5*base*height) or a trapezoid (which can typically be transformed into a rectangle and a triangle or the area can be computed as 0.5*(h1 + h2)*base).
a. Area = 0.5*base*height = 0.50*(5.0 s)*(20.0 m/s) = 50. m
b. Area = A1 + A2 (see diagram) = (5.0 s)*(5.0 m/s) + 0.50*(5.0 s)*(10.0 m/s) = 25 m + 25 m = 50. m
c. Area = A1 + A2 + A3 + A4 (see diagram)
Area = 0.5*(5.0 s)*(20.0 m/s) + (5.0 s)*(20.0 m/s) + (5.0 s)*(20.0 m/s) + 0.5*(5.0 s)*(10.0 m/s)
Area = 50 m + 100 m + 100 m + 25 m = 275 m
43.
a. If the speed and direction of an object is constant, then the acceleration is 0 m/s2.
b. The acceleration is the velocity change per time ratio:
c. The acceleration is the velocity change per time ratio:
14.3 mi/hr/s * (1.0 m/s) / (2.24 mi/hr) = 6.38 m/s2.
d. The acceleration value can also be calculated using kinematic equations if three other kinematic quantities are known. In this case, the know information is: vo = 33.4 m/s; vf = 18.9 m/s; and d = 109 m. Using the equation vf2 = vo2 + 2*a*d, the acceleration can be computed.
44.
a. Since this is a round-trip journey, the overall displacement is 0 m.
b. Since the velocity is constant, the displacement can be found by multiplying the velocity by the time.
c. A displacement value can also be calculated using kinematic equations if three other kinematic quantities are known. In this case, the know information is: vo = 38.1 m/s; vf = 17.6 m/s; and a = -4.35 m/s/s. Using the equation vf2 = vo2 + 2*a*d, the displacement can be computed.
d. A displacement value can be calculated using other kinematic equations if a different set of kinematic quantities is known. Here we know that: vo = 0.0 m/s; t = 12.1 s; and a = 3.67 m/s/s. Using the equation d = vo* t + 0.5*a*t2, the displacement can be computed.
e. Here the displacement value is calculated using the same kinematic equation. We know that: vo = 12.2 m/s; t = 17.0 s; and a = 1.88 m/s/s. Using the equation d = vo* t + 0.5*a*t2, the displacement can be computed.
45.
Answer: 0. 67 m/s2
Like a lot of physics word problems, there is more than one path to the final answer. In all such problems, the solution involves thought and good problem-solving strategies (draw a picture, list what you know, list pertinent equations, etc.).
The tortoise, moving at a constant speed, will cover the 1200 m in a time of:
The hare will sleep for 6.5 hrs (23400 s) before starting, and so will have only 0.1666... hrs (600 s) to accelerate to the finish line. So the acceleration of the hare can be determine using a kinematic equation. The know information about the hare's motion is: t = 600 s; d = 1200 m; vo = 0 m/s. The best equation is d = vo* t + 0.5*a*t2. The vo* t term cancels and the equation can be algebraically rearranged and solved for a:
46.
Answer: 14.0 s
(As mentioned in the previous problem ...) Like a lot of physics word problems, there is more than one path to the final answer. In all such problems, the solution involves thought and good problem-solving strategies (draw a picture, list what you know, list pertinent equations, etc.).
Here the gold car travels with a constant speed for a time of t seconds (where t is the total time of travel for both cars). The distance traveled by the gold car is given by the kinematic equation d = vo* t + 0.5*a*t2. The second term cancels and the distance can be expressed as
dgold = 12.0* t
For the green car, there is an accelerated period and then a constant speed period. The distance traveled during the accelerated period (d1green) is found from the same kinematic equation. For the green car, the first term cancels and the distance is
d1green = 108.9 m
Once the green car has accelerated for 11 seconds, it maintains a constant speed for the remaining time, given by the expression t - 11 s. The speed at which the green car is traveling during this time can be computed using the equation:
The distance traveled by the green car during this constant speed portion of its motion (d2green) can be computed using the kinematic equation. d = vo* t + 0.5*a*t2. The second term cancels and the distance can be expressed as
d2green = 19.8*t - 217.8
So the total distance traveled by the green car is given by the expression:
dgreen = 19.8*t - 108.9
When the green car catches up to the gold car their distance traveled will be the same. So the time t can be determined by setting the two expressions for distance equal to each other and solving for t.
108.9 = 7.80*t
t = (108.9) / (7.80)
t = 13.96 s = 14.0 s
47.
Answer: 0.486 s
Ima's total distance traveled (62.0 m) can be broken into two segments - a reaction distance (drxn) and a braking distance (dbraking). The reaction distance is the distance Ima moves prior to braking; she will move at constant speed during this time of trxn. The braking distance is the distance which Ima moves when her foot is on the brake and she decelerates from 28.0 m/s to 0.0 m/s. The braking distance can be computed first using the following kinematic equation: vf2 = vo2 + 2*a*d. The known information for this braking period is: vo = 28 m/s; vf = 0 m/s; and a = -8.10 m/s/s. The substitutions and solution are shown below.
Since Ima's car requires 48.40 m to brake, she can travel a maximum of 13.6 m during the reaction period. The relationship between reaction time, speed and reaction distance is given by the equation
Substituting 13.6 m for drxn and 28.0 m/s for v, the reaction time can be computed:
48.
This problem can be approached by either the use of a velocity-time graph or the use of kinematic equations (or a combination of each). Whatever the approach, it is imperative to break the multistage motion up into its three different acceleration periods. The use of kinematic equations is only appropriate for constant acceleration periods. For this reason, the complex motion must be broken up into time periods during which the acceleration is constant. These three time periods can be seen on the velocity-time graph by three lines of distinctly different slope. The diagram at the right provides a depiction of the motion; strategic points are labeled. These points will be referred to in the solutions below. The velocity-time plot below will be used throughout the solution; note that the same strategic points are labeled on the plot.
a. The maximum speed occurs after the second stage or acceleration period (point C). After this time, the upward-moving rocket begins to slow down as gravity becomes the sole force acting upon it. To determine this speed (vc), the kinematic equation vf = vo + a*t will be used twice - once for each acceleration period.
Second Stage: vC = vB + a*t = 24.3 m/s + (2.98 m/s/s) * (5.90 s) = 41.9 m/s
b. The maximum altitude occurs at point D, sometime after the second stage has ceased and the rocket finally runs out of steam. The velocity at this point is 0 m/s (it is at the peak of its trajectory). The altitude at this point is the cumulative distance traveled from t = 0 s to t = tD. This distance is the distance for the first stage, the second stage and the deceleration period (C to D). These distances correspond to the area on the v-t graph; they are labeled A1, A2, and A3 on the graph. They are calculated and summed below.
A2 = b*h + 0.5*b*h (a triangle on top of a square)
A2 = (24.3 m/s) * (5.9 s) + 0.5 * (41.9 m/s - 24.3 m/s) * (5.9 s) = 196.65 m
It will be necessary to know the time from point C to point D in order to determine A3. This time can be determined using the kinematic equation vf = vo + a*t for which vf = 0 m/s and vo = 41.9 m/s and a = -9.8 m/s/s.
0 m/s = 41.9 m/s + (-9.8 m/s/s) * t
t = 4.28 s
Now A3 can be determined using the v-t graph. The area is a triangle and is calculated as
The maximum altitude is the sum of the three distances (areas)
c. When the rocket reaches point D, the time is 17.0 seconds. The altitude at 20.0 seconds will be the 369 meters risen above the launch pad from point A to point D minus the distance fallen from the peak from 17.0 to 20.0 seconds. This distance would be represented by a negative area on the velocity-time graph. The area is a triangle and can be computed if the velocity at 20 seconds is known. It can be calculated using a kinematic equation and then used to determine the area of a triangle. Alternatively, a kinematic equation can be used to determine the distance fallen during these 3.0 seconds. The work is shown below:
The altitude at 20 seconds is therefore the ~369 m risen in the first 17 seconds minus the ~44 m fallen in the next 3 seconds. The answer is 325 m.
d. The rocket rises 369 m in the first 17.0 seconds. In the time subsequent of this, the rocket must fall 369 meters. The time to fall 369 m can be found from the same kinematic equation used in part c.
-369 m = 0.5 * (-9.8 m/s/s) * t2
t = 8.68 seconds
This time can be added onto the 17.0 seconds to determine the time at which the rocket lands: 25.7 seconds.
49.
Answer: 4.04 m/s
Both swimmers swim the same distance (50 m) at constant speed. Swimmer A (who was just arbitrarily named) gets a 0.450 second head start. So swimmer B must travel faster in order to finish the race in less time than swimmer A. First, the time required of swimmer A to complete the 50l0 m at 3.90 m/s can be computed. The time is
Thus, swimmer B must finish the same 50.0 m in 12.37 seconds (12.82s - 0.45 s). So the speed of swimmer B can be computed as
50.
Answer: -24 m/s/s
This problem can be approached by first determining the distance over which the dragster decelerates. This distance will be less than 80. meters by an amount equal to the distance which the dragster coasts after crossing the finish line. See diagram.
The distance traveled by the dragster prior to braking is 100 m plus the coasting distance. The coasting distance can be determined if the speed of the dragster at the end of 100 m is determined. So first, a kinematic equation will be used to determine the speed and then the coasting distance will be computed.
Using the equation vf2 = vo2 + 2*a*d, the speed after 100 m can be determined. This substitution and solution is shown below.
vf = 51.4 m/s
Coasting at 51.38 m/s for 0.500 s will lead to a distance traveled of 25.7 m.
Once the coasting period is over, there is a short distance left to decelerate to a stop. This distance is
Now the same kinematic equation can be used to determine the deceleration rate during the last 54 m of the track. Known information is: vo = 51.4 m/s; vf = 0 m/s; and d = 54 m. Using the equation vf2 = vo2 + 2*a*d, the acceleration can be computed.
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