Problem#1
An electron with an initial velocity v0 = 1.50 x 105 m/s enters a region of length L = 1.00 cm where it is electrically accelerated (Fig. 1). It emerges with v = 5.70 x 106 m/s. What is its acceleration, assumed constant? |
| Fig.1 |
Answer:
Since the problem involves constant acceleration, the motion of the electron can be readily analyzed using the equations v = v0 + at, x = x0 + v0t + ½ at2 and v2 = v02 + 2a(x – x0), then
a = (v2 – v02)/2∆x
a = [(5.70 x 106 m/s)2 – (1.50 x 106 m/s)2]/[2(0.01 m)] = 1.62 x 1015 m/s2
Problem#2
An electric vehicle starts from rest and accelerates at a rate of 2.0 m/s2 in a straight line until it reaches a speed of 20 m/s. The vehicle then slows at a constant rate of 1.0 m/s2 until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle travel from start to stop?
Answer:
We separate the motion into two parts, and take the direction of motion to be positive. In part 1, the vehicle accelerates from rest to its highest speed; we are given
v0 = 0; v = 20 m/s and a = 2.0 m/s2
In part 2, the vehicle decelerates from its highest speed to a halt; we are given
v0 = 20 m/s; v = 0 and a = –1.0 m/s2 (negative because the acceleration vector points opposite to the direction of motion).
(a) we find t1 (the duration of part 1) from
v = v0 + at
In this way,
20 = 0 + 2t1
yields t1 = 10 s
We obtain the duration t2 of part 2 from the same equation.
Thus, 0 = 20 + (–1.0)t2
leads to t2 = 20 s,
and the total is t = t1 + t2 = 30 s.
(b) For part 1, taking x0 = 0, we use the equation
v2 = v02 + 2a(x – x0)
and find
x – x0 = (v2 – v02)/2a
x – 0 = [(20 m/s)2 – 0]/[2(2.0 m/s2)]
x = 100 m
This position is then the initial position for part 2, so that when the same equation is used in part 2 we obtain
x – 100 = (v2 – v02)/2a = [0 – (20 m/s)2]/[2(1 – 1.0 m/s2)]
x = 300 m
Thus, the final position is x = 300 m. That this is also the total distance traveled should be evident (the vehicle did not "backtrack" or reverse its direction of motion).
Problem#3
An electron has a constant acceleration of +3.2 m/s2. At a certain instant its velocity is +9.6 m/s. What is its velocity (a) 2.5 s earlier and (b) 2.5 s later?
Answer:
We use v = v0 + at, with t = 0 as the instant when the velocity equals +9.6 m/s.
(a) Since we wish to calculate the velocity for a time before t = 0, we set t = –2.5 s. Thus,
v = v0 + at, gives
v = 9.6 m/s + (3.2 m/s2)(–2.5 s) = 1.6 m/s
(b) Now, t = +2.5 s and we find
v = 9.6 m/s + (3.2 m/s2)(+2.5 s) = 18 m/s
Problem#4
A certain elevator cab has a total run of 190 m and a maximum speed of 305 m/min, and it accelerates from rest and then back to rest at 1.22 m/s2. (a) How far does the cab move while accelerating to full speed from rest? (b) How long does it take to make the nonstop 190 m run, starting and ending at rest?
Answer:
We assume the periods of acceleration (duration t1) and deceleration (duration t) are periods of constant a so that Table 2-1 can be used. Taking the direction of motion to be +x then a1 = +1.22 m/s2 and a2 = –1.22 m/s2 . We use SI units so the velocity at t = t1 is v = 305/60 = 5.08 m/s.
(a) We denote Δx as the distance moved during t1, and use
v2 = v02 + 2a(x – x0)
∆x = (v2 – v02)/2a = [(5.08 m/s)2 – 0]/[2(1.22 m/s2)] = 10.59 m
(b) Using Eq. v = v0 + at, we have
t1 = (v – v0)/a = (5.08 m/s)/(1.22 m/s2) = 4.17 s
The deceleration time t2 turns out to be the same so that t1 + t2 = 8.33 s. The distances traveled during t1 and t2 are the same so that they total to
= 2(10.59 m) = 21.18 m
This implies that for a distance of
= 190 m – 21.18 m = 168.82 m,
the elevator is traveling at constant velocity. This time of constant velocity motion is
t3 = ∆x/v = 168.82 m/(5.08 m/s) = 33.21 s
Therefore, the total time is 8.33 s + 33.21 s ≈ 41.5 s.
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