Problem#1
The position of a pinewood derby car was observed at various times; the results are summarized in the following table.
(a) Use the data in Problem 1 to construct a smooth graph of position versus time. (b) By constructing tangents to the x(t) curve, find the instantaneous velocity of the car at several instants. (c) Plot the instantaneous velocity versus time and, from this, determine the average acceleration of the car. (d) What was the initial velocity of the car?
Answer:
(a) Use the data in to construct a smooth graph of position versus time in Fig.1a.
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| Fig.1 |
(b) At t = 5.0 s, the slope is v = 58 m/2.5 s ≅ 23 m/s
At t = 4.0 s, the slope is v = 54 m/3.0 s ≅ 18 m/s
At t = 3.0 s, the slope is v = 49 m/3.4 s ≅ 14 m/s
At t = 2.0 s, the slope is v = 36 m/4.0 s ≅ 9 m/s
(c) Plot the instantaneous velocity versus time in Fig. 3b and, from this, and the average acceleration of the car is
aavg = ∆v/∆t = 23 m/s/5.0 s = 4.6 m/s2
(d) Initial velocity of the car was zero
Problem#2
A position-time graph for a particle moving along the x axis is shown in Figure 2. (a) Find the average velocity in the time interval t = 1.50 s to t = 4.00 s. (b) Determine the instantaneous velocity at t = 2.00 s by measuring the slope of the tangent line shown in the graph. (c) At what value of t is the velocity zero?
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| Fig.2 |
Answer:
For average velocity, we fi nd the slope of a secant line running across the graph between the 1.5-s and 4-s points. Then for instantaneous velocities we think of slopes of tangent lines, which means the slope of the graph itself at a point.
(a) From the graph: At t1 = 1.5 s, x = x1 = 8.0 m and At t2 = 4.0 s, x = x2 = 2.0 m
Therefore, vavg = ∆x/∆t = (2.0 m – 8.0 m)/(4.0 s – 1.5 s) = –2.4 m/s.
(b) Choose two points along a line which is tangent to the curve at t = 2.0 s. We will use the two points (ti = 0.0 s, and xi = 13.0 m) and (tf = 3.5 s, xf = 0.0 m). Instantaneous velocity equals the slope of the tangent line,
So, vx = = (0.0 m – 13.0 m)/(3.5 s – 0.0 s) = –3.7 m/s.
The negative sign shows that the direction of vx is along the negative x direction. ■
(c) The velocity will be zero when the slope of the tangent line is zero. This occurs for the point on the graph where x has its minimum value. Therefore, v = 0 at t = 4.0 s.
Problem#3
Find the instantaneous velocity of the particle described in Figure 3 at the following times: (a) t = 1.0 s, (b) t = 3.0 s, (c) t = 4.5 s, and (d) t = 7.5 s.
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| Fig.3 |
Answer:
(a) the instantaneous velocity of the particle at the following times t = 1.0 s is
v = (5 m – 0)/(1 s – 0 s) = 5 m/s
(b) the instantaneous velocity of the particle at the following times t = 3.0 s is
v = (5 m – 10 m)/(4 s – 2 s) = –2.5 m/s
(c) the instantaneous velocity of the particle at the following times t = 4.5 s is
v = (5 m – 5 m)/(5 s – 4 s) = 0
(d) the instantaneous velocity of the particle at the following times t = 4.5 s is
v = (0 – (–5 m))/(8 s – 7 s) = +5 m/s
Problem#4
A hare and a tortoise compete in a race over a course 1.00 km long. The tortoise crawls straight and steadily at its maximum speed of 0.200 m/s toward the finish line. The hare runs at its maximum speed of 8.00 m/s toward the goal for 0.800 km and then stops to tease the tortoise. How close to the goal can the hare let the tortoise approach before resuming the race, which the tortoise wins in a photo finish? Assume that, when moving, both animals move steadily at their respective maximum speeds.
Answer:
Once it resumes the race, the hare will run for a time of
t = ∆x/vx = (1000 m – 800 m)/(8 m/s) = 25 s
In this time, the tortoise can crawl a distance
xf −xi = 0.2 m/s x 25 s= 5.00 m
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