Problem#1
Jules Verne in 1865 suggested sending people to the Moon by firing a space capsule from a 220-m-long cannon with a launch speed of 10.97 km/s. What would have been the unrealistically large acceleration experienced by the space travelers during launch? Compare your answer with the free-fall acceleration 9.80 m/s2.Answer:
From v2 = v02 + 2a(x – x0)
a = (v2 – v02)/2∆x
a = [(10.97 x 103 m/s)2 – 0]/[2(220 m)] = 2.74 x 105 m/s2
which is
a = 2.79 x 104 g
Problem#2
An object moving with uniform acceleration has a velocity of 12.0 cm/s in the positive x direction when its x coordinate is 3.00 cm. If its x coordinate 2.00 s later is +5.00 cm, what is its acceleration?
Answer:
Given v0 = 12.0 cm/s when x0 = 3.00 cm (t = 0), and at t = 2.00 s, x = −5.00 cm,
We use
x = x0 + v0t + ½ at2
−5.00 x 10-2 m = −3.00 x 10-2 m + (0.12 m/s)(2.00 s) + ½ a(2.00 s)
which yields
a = –0.16 m/s2
Problem#3
A speedboat moving at 30.0 m/s approaches a no-wake buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of -3.50 m/s2 by reducing the throttle. (a) How long does it take the boat to reach the buoy? (b) What is the velocity of the boat when it reaches the buoy?
Answer:
(a) Choose the initial point where the pilot reduces the throttle and the final point where the boat passes the buoy:
Given: x0 = 0 , x =100 m , vx0 = 30 m/s, vxf =?, ax =−3.5 m/s2 , t = ?
We use
x – x0 = v0xt + ½ axt2
100 m – 0 = (30 m/s)t + ½ (–3.5 m/s2)t2
(1.75 m/s2)t2 – (30 m/s)t + 100 m = 0
We use the quadratic formula:
t = [–b ± (b2 – 4ac)1/2]/2a
= {(–30 m/s) ± [(30 m/s)2 – 4(1.75 m/s2)(100 m)]}/(2 x 1.75 m/s2)
t = 12.6 s or 4.53 s
The smaller value is the physical answer. If the boat kept moving with the same acceleration, it would stop and move backward, then gain speed, and pass the buoy again at 12.6 s.
(b) the velocity of the boat when it reaches the buoy is
vx = v0x + at = 30 m/s + (–3.5 m/s2)(4.53 s)
vx = 14.1 m/s
Problem#4
A particle moves along the x axis. Its position is given by the equation x = 2 + 3t – 4t2 with x in meters and t in seconds. Determine (a) its position when it changes direction and (b) its velocity when it returns to the position it had at t = 0.
Answer:
(a) Compare the position equation x = 2 + 3t – 4t2 to the general form
x = x0 + v0t + ½ at2
to recognize that x0 = 2.00 m, v0 = 3 00 m/s, and a = −8.00 m/s2. The velocity equation,
v = v0 + at is then
v = 3.00 m/s – (8.00 m/s2)t
The particle changes direction when v = 0 , which occurs at t = 3/8 s. The position at this time is:
x = 2.00 m + (3.00 m/s)(3/8 s) – (4.00 m/s2)(3/8 s)2 = 2.56 m
(b) From
x = x0 + v0t + ½ at2
observe that when x = x0 , the time is given by
t = −2v0/a
Thus, when the particle returns to its initial position, the time is
t = −2(3.00 m/s)/(–8.00 m/s2) = 0.75 s
and the velocity is
v = 3.00 m/s – (8.00 m/s2)(0.75 s)
v = –3.00 m/s
Post a Comment for "One-Dimensional Motion with Constant Acceleration Problems and Solutions"