Problem#1
A 50.0-g superball traveling at 25.0 m/s bounces off a brick wall and rebounds at 22.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.50 ms, what is the magnitude of the average acceleration of the ball during this time interval? (Note: 1 ms = 10-3 s.)Answer:
Choose the positive direction to be the outward direction, perpendicular to the wall
a = ∆v/∆t = [(22.0 m/s) – (–25.0 m/s)]/(3.50 x 10-3 s)
a = 1.34 x 104 m/s2
Problem#2
A particle starts from rest and accelerates as shown in Figure 1. Determine (a) the particle’s speed at t = 10.0 s and at t = 20.0 s, and (b) the distance traveled in the first 20.0 s.
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| Fig.1 |
Answer:
(a) Acceleration is constant over the first ten seconds, so at the end,
a = ∆v/∆t = (vf – vi)/t
2.0 m/s2 = (vf – 0)/10 s
vf = 20.0 m/s
Then a = 0 so v is constant from t =10 0. s to t =15 0. s. And over the last five seconds the velocity changes to
a = ∆v/∆t = (vf – vi)/t
2.0 m/s2 = (vf – 20.0 m/s)/5 s
vf = 5.00 m/s
(b) In the first ten seconds,
xf = xi + vit + ½ at2
xf = 0 + 0 + ½(2.00 m/s2)(10.0 s)2 = 100 m
Over the next five seconds the position changes to
xf = xi + vit + ½ at2
xf = 100 + (20.0 m/s)(5.00 s) + 0 = 200 m
And at t = 20.0 s,
xf = xi + vit + ½ at2
xf = 200 + (20.0 m/s)(5.00 s) + ½ (–3 m/s2)(5.00 s)2 = 262 m
Problem#3
Secretariat won the Kentucky Derby with times for successive quarter-mile segments of 25.2 s, 24.0 s, 23.8 s, and 23.0 s. (a) Find his average speed during each quarter-mile segment. (b) Assuming that Secretariat’s instantaneous speed at the finish line was the same as the average speed during the final quarter mile, find his average acceleration for the entire race. (Horses in the Derby start from rest.)
Answer:
The average speed during a time interval ∆t is
Savg = distance traveled/∆t
During the first quarter mile segment, Secretariat’s average speed was
Savg, 1 = 0.250 mi/25.2 s = 1320 ft/25.2 s = 52.4 ft/s = 35.6 mi/s
During the second quarter mile segment,
Savg, 2 = 1320 ft/24 s = 55.0 ft/s = 37.4 mi/s
For the third quarter mile of the race,
Savg, 3 = 1320 ft/23.8 s = 55.5 ft/s = 37.7 mi/s
and during the final quarter mile,
Savg, 4 = 1320 ft/23 s = 57.4 ft/s = 39 mi/s
(b) Assuming that vf = vavg,4 and recognizing that vi = 0 , the average acceleration during the race was
aavg = (57.4 ft/s – 0)/(25.2 s + 24.0 s + 23.8 s + 23.0 s) = 0.598 m/s2
Problem#4
A velocity–time graph for an object moving along the x axis is shown in Figure 2. (a) Plot a graph of the acceleration versus time. (b) Determine the average acceleration of the object in the time intervals t = 5.00 s to t = 15.0 s and t = 0 to t = 20.0 s.
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| Fig.2 |
Answer:
(a) Acceleration is the slope of the graph of v vs t.
For 0 < t < 5.00 s, a = 0
For 15.0 s < t < 20.0 s, a = 0
For 5.0 s < t < 15.0 s,
a = ∆v/∆t = (8.00 m/s – (–8.00 m/s))/(15.0 s – 5.0 s) = 1.60 m/s2
We can plot a(t) as shown Fig.3.
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| Fig.3 |
(b) the average acceleration of the object in the time intervals t = 5.00 s to t = 15.0 s and t = 0 to t = 20.0 s.
For the time intervals t = 5.00 s to t = 15.0 s,
aavg = ∆v/∆t = (8.00 m/s – (–8.00 m/s))/(15.0 s – 5.0 s) = 1.60 m/s2
For the time intervals t = 0 s to t = 20.0 s,
aavg = ∆v/∆t = (8.00 m/s – (–8.00 m/s))/(20.0 s – 0) = 0.80 m/s2
Problem#5
A particle moves along the x axis according to the equation x = 2.00 + 3.00t – 1.00t2, where x is in meters and t is in seconds. At t = 3.00 s, find (a) the position of the particle, (b) its velocity, and (c) its acceleration.
Answer:
the equation x = 2.00 + 3.00t – 1.00t2, then
vx = dx/dt = 3.00 – 2.00t and
ax = dvx/dt = –2.00 m/s2
at t = 3.00 s,
(a) x = 2.00 + (3.00)(3.00) – 1.00(3.00)2 = 2.00 m
(b) v = 3.00 – 2.00(3.00) = –3.00 m/s
(c) ax = –2.00 m/s2
Problem#5
An object moves along the x axis according to the equation x(t) = (3.00t2 – 2.00t + 3.00) m. Determine (a) the average velocity between t = 2.00 s and t = 3.00 s, (b) the instantaneous speed at t = 2.00 s and at t = 3.00 s, (c) the average acceleration between t = 2.00 s and t = 3.00 s, and (d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s.
Answer:
the equation x(t) = (3.00t2 – 2.00t + 3.00) m
At t = 2.00 s, x = 3.00(2.00)2 – 2.00(2.00) + 3.00 = 11.0 m
At t = 3.00 s, x = 3.00(3.00)2 – 2.00(3.00) + 3.00 = 24.0 m
(a) the average velocity between t = 2.00 s and t = 3.00 s is
vavg = ∆x/∆t = (24.0 m – 11.0 m)/(3.00 s – 2.00 s) = 13.0 m/s
(b) the instantaneous speed at t = 2.00 s and at t = 3.00 s, is
v = dx/dt = 6.00t – 2.00
at t = 2.00 s; v = 6.00(2.00) – 2.00 = 10.0 m/s
at t = 3.00 s; v = 6.00(3.00) – 2.00 = 16.0 m/s
(c) the average acceleration between t = 2.00 s and t = 3.00 s is
aavg = ∆v/∆t = [(16.0 m/s) – (10.0 m/s)]/(3.00 s – 2.00 s) = 6.00 m/s2
(d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s, is
a = dv/dt = 6.00 m/s2 (This includes both t = 2.00 s and t = 3.00 s)
Problem#6
Figure 4 shows a graph of vx versus t for the motion of a motorcyclist as he starts from rest and moves along the road in a straight line. (a) Find the average acceleration for the time interval t = 0 to t = 6.00 s. (b) Estimate the time at which the acceleration has its greatest positive value and the value of the acceleration at that instant. (c) When is the acceleration zero? (d) Estimate the maximum negative value of the acceleration and the time at which it occurs.
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| Fig.4 |
Answer:
(a) the average acceleration for the time interval t = 0 to t = 6.00 s is
aavg = ∆v/∆t = (8.00 m/s)/(6.00 s) = 1.33 m/s2
(b) maximum positive acceleration is at t = 3 s, and is approximately 2 m/s2.
(c) the acceleration zero (a = 0) at t = 6.0 s and also for t > 10 s.
(d) maximum negative acceleration is at t = 8 s, and is approximately –1.5 m/s2.
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