Problem#1
Suppose a rocket ship in deep space moves with constant acceleration equal to 9.8 m/s2, which gives the illusion of normal gravity during the flight. (a) If it starts from rest, how long will it take to acquire a speed one-tenth that of light, which travels at 3.0 x 108 m/s? (b) How far will it travel in so doing?
Answer:
(a) We solve v = v0 + at for the time:
t = (v – v0)/a = [(0.1 x 3.0 x 108 m/s) – 0]/(9.8 m/s2) = 3.1 x 106 s
which is equivalent to 1.2 months.
(b) We evaluate x = x0 + v0t + ½ at2, with x0 = 0, v0 = 0. The result is
x = ½ (9.8 m/s2)(3.1 x 106 s) = 4.6 x 1013 m
Note that in solving parts (a) and (b), we did not use the equation
v2 = v02 + 2a∆x
This equation can be employed for consistency check. The final velocity based on this equation is
v2 = 2(9.8 m/s2)(4.6 x 1013 m)
v = 3.0 x 107 m/s
Problem#2
A car traveling 56.0 km/h is 24.0 m from a barrier when the driver slams on the brakes.The car hits the barrier 2.00 s later. (a) What is the magnitude of the car’s constant acceleration before impact? (b) How fast is the car traveling at impact?
Answer:
We take x0 = 0, and solve
x = v0t + ½ at2
Substituting x = 24.0 m, v0 = 56.0 km/h = 15.55 m/s and t = 2.00 s, we find
24.0 m = (15.55 m/s)(2.00 s) + ½ a(2.00 s)2
a = –7.1 m/2 s2 = –3.56 m/s2
or |a| 3.56 m/s2. The negative sign indicates that the acceleration is opposite to the direction of motion of the car. The car is slowing down.
(b) We evaluate v = v0 + at as follows:
v = 15.55 m/s + (–3.56 m/s2)(2.00 s) = 8.43 m/s
which can also be converted to 30.3 km/h
Problem#3
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| Fig.1 |
Figure 1a shows a red car and a green car that move toward each other. Figure 1b is a graph of their motion, showing the positions xg0 = 270 m and xr0 = -35.0 m at time t = 0. The green car has a constant speed of 20.0 m/s and the red car begins from rest.What is the acceleration magnitude of the red car?
Answer:
The positions of the cars as a function of time are given by
xr(t) = xr0 + ½ art2 = (–35.0 m) + ½ art2
xg(t) = xg0 + vgt = (270.0 m) – (20.0 m/s)t
where we have substituted the velocity and not the speed for the green car. The two cars pass each other at t = 12.0 s when the graphed lines cross. This implies that
(–35.0 m) + ½ ar(20.0 s)2 = (270.0 m) – (20.0 m/s)(20.0 s)
ar = 0.90 m/s2
Problem#4
Figure 2 depicts the motion of a particle moving along an x axis with a constant acceleration. The figure’s vertical scaling is set by xs = 6.0 m.What are the (a) magnitude and (b) direction of the particle’s acceleration?
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| Fig.2 |
Answer:
(a) From the figure, we see that x0 = –2.0 m. We can apply
x – x0 = v0t + ½ at2
with t = 1.0 s, and then again with t = 2.0 s. This yields two equations for the two unknowns, v0 and a:
0 – (–2.0 m) = v0(1.0 s) + ½ a(1.0 s)2
6.0 m – (–2.0 m) = v0(2.0 s) + ½ a(2.0 s)2
Solving these simultaneous equations yields the results v0 = 0 and a = 4.0 m/s2.
(b) The fact that the answer is positive tells us that the acceleration vector points in the +x direction.
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