1. The supermarket tabloids report that a (rather clumsy) person fell 137 ft fro
ID: 1647035 • Letter: 1
Question
1. The supermarket tabloids report that a (rather clumsy) person fell 137 ft from a tall building to the top of a metal ventilator box. The metal ventilator box was crushed 18.7 inches and he suffered only minor injuries. In the following questions use a coordinate system with down being positive. (a) What is the person’s speed (m/s) just before colliding with the ventilator box? (b) What is the person’s acceleration (m/s2) while crushing the box? Assume constant acceleration. (c) How long did it take (ms) for the person to come to a stop after first contacting the box?
2. Your good friend’s car (mentioned above) ignores a stop sign and enters an intersection with a speed v0 of 26.9 m/s. He then continues with a constant velocity. A motorcycle traffic patrol officer is parked at the intersection. He discards his hot cocoa and doughnut and begins accelerating (at the instant the car passes him) with a constant acceleration of 3.29 m/s2. (a) How long (s) does it take the officer to catch the car? (b) How far (m) has the officer traveled when he catches up to the car? (c) How fast (m/s) is the officer going when he catches up to the car?
3. A particle moves in such a way that its position z, in meters, is given as a function of t, in seconds, by the equation z = c1t2 c2t3 where c1 = 2.12 m/s2 and c2 = 3.46 m/s3. (a) The particle is at z = 0 twice. What is the time (s) of the second occurrence of z = 0? (b) What is the displacement z (m) of the particle between the times t = 2.00 s and t = 5.00 s? (c) What is the velocity (m/s) of the particle at t = 0.657 s? (d) The velocity is also zero twice. What is the time (s) of the second occurrence? (e) What is the acceleration (m/s2) at t = 0.657 s? (f) When is the acceleration equal to zero (s)? (g) When the acceleration is zero, what is the particle’s position (m)?
Explanation / Answer
1. Conservation of Energy is easier than Kinematics for this
(a) mgh = 0.5*m*v^2
So, mass doesnt matter
v = sqrt(2gh) = sqrt(2*32.2*137) = 93.93 ft/s
(b) The persons acceleration crushing box is found same way
v^2 = v0^2+2a(x-x0) = 0
v^2 = 0 = 93.93^2+2(a)(1.55 ft)
a = 93.93^2/(2*1.55)
a = 2846.08 ft/s^2
(c) The time to stop comes from
x = x0+v0*t+(1/2)at^2
where, x = 0
x0 = 18.7
v0 = 93.93 ft/s, and
a = 2846.08 ft/s^2
Use the quadratic formula to solve
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