Problem Set 2.11: | 60
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| 73
| Go
up** - by Chris Murray, 9/2001 and Kelly
Intile 12/03**

**60. A person jumps from a fourth-story window 15.0 m above
a firefighter's safety net. The survivor stretches the net 1.0 m before
coming to rest, Fig 2-35. (a) What was the average deceleration
experienced by the survivor when slowed to rest by then net? (b) What would you
do to make it "safer" (that is, generate a smaller deceleration):
would you stiffen or loosen the net? Explain.**

<Fig 2-35>

(a) First we need to find the final velocity when the person was falling and the net caught them. Use SUVAT.

s = - 15 m. u = 0 m/s a = - 9.8 m/s/s (gravity) v = ???

v

^{2}= u^{2}+ 2as v² = 0² + 2(-9.8)(.15)Do the math and you should get that v = ±Ö294. Common sense says that since the person is falling, their velocity should be negative, so v = -17.14643

Now that we have the velocity, we can solve for the acceleration of the person when they're in the net. SUVAT again.

s = 1.0 m. u = -17.14643 v = 0 m/s (the net slows them to rest) a = ???

v

^{2}= u^{2}+ 2as 0² = (.17.14643)² + 2*1.0*aDo math and you should get that a = -147 m/s/s, but since they asked for deceleration (meaning negative acceleration) your final answer should be 147 m/s/s or 150 m/s/s with sig figs.

(b) To make the deceleration less, you would want to loosen the net so that there would be a greater displacement, allowing the person a greater distance over which to slow down.

**61. The acceleration due to gravity on the Moon is about
one sixth what it is on Earth. If an object is thrown vertically upward on
the Moon, how many times higher will it go than it would on Earth, assuming the
same initial velocity?**

Use any formula with acceleration and displacement in it and see what happens.

v

^{2}= u^{2}+ 2as Pretend that initial velocity is 10 m/s and final velocity is 0 m/s (the object reaches its peak). First work out this equation using acceleration due to gravity on Earth (a = -9.8 m/s/s). Displacement ends up being 5.102 m.Now do the same formula with the same initial and final velocities, but use acceleration on the moon (-9.8/6 = -1.633333). You will find that the displacement on the moon is 30.612, which coincidentally is exactly

six times greaterthan the displacement on Earth.

**62. A person who is properly constrained by an
over-the-shoulder seat belt has a good chance of surviving a car collision if
the deceleration does not exceed 30 "g's" (1.00 g = 9.80 m/s/s).
Assuming uniform deceleration of this value, calculate the distance over which
the front end of the car must be designed to collapse if a crash being the car
to rest from 100 km/h. **

The first thing to do is to get rid of km/h because we need m/s. Divide 100 km/h by 3.6 and you get 27.77778 m/s. We also need to know what exactly 30 "g's" is. 30*9.8 = 294 m/s/s. Now we can use SUVAT to solve for displacement.

v

^{2}= u^{2}+ 2as v = 0 (car crashes to rest) u = 27.77778 m/s a = 294 s = ???Do the math and you will find that displacement needs to be 1.132 meters, of 1.3 m. with sig figs.

(Table of contents)

**63. A race car driver must average 200.0 km/h over the
course of a time trial lasting ten laps. If the first nine laps were done
at 199.0 km/h, what average speed must be maintained for the last lap?**

We only need to use v = s/t for this problem, but change it to t = s/v. The total time is (total distance)/(average velocity). We know that the total distance is 10 laps, and we will use

xto represent how many kilometers are in a lap. We also know that the driver must maintain an average of 200 km/h over the ten laps.Total time = (10

x)/200We can also apply this same formula to the first nine laps that he has already completed. (Average speed for the first nine laps is 199 km/h)

Time so far = (9

x)/199So, logically, the time he has left is total time - time so far.......(10

x)/200 - (9x)/199. This gets a little ugly, but find a common denominator and subtract these fractions and you should end up with time = (19x)/3980.Now, we know how much time the driver has left and we know how far he has to go (one lap, or

xkm.), so we can use the original equation v = s/t to figure out what his velocity must be.velocity = distance/time =

x/(19x/3980). *Remember that we're not solving forx- they end up cancelling outSimplify and you get that velocity = 209.5 m/s

(Table of contents)

**65. A first stone is dropped from the roof of a
building 2.00 s after that a second stone is thrown straight down with an
initial speed of 30.0 m/s, and it is observed that the two stones land at the
same time. (a) How long did it take the first stone to reach the
ground? (b) How high is the building? (c) What are the speeds of the two
stones just before they hit the ground? **

a) The stone thrown later overtakes the dropped stone just as it reaches the ground. The formula to use is

s = ut +^{1}/_{2}at^{2}- but therein lies the rub - we have two rocks - one dropped and one thrown down 2.00 seconds later. We have two equations of motion:

s =^{1}/_{2}at^{2}- for the dropped rock (u = 0)

s = ut +^{1}/_{2}at^{2}- for the thrown down rock - but it is a bit more complicated than that. for the thrown rock it really is

s = u(t-2.00) +^{1}/_{2}a(t-2.00)^{2}- because it is thrown 2.00 seconds later. Plugging in for the initial velocity and acceleration (making down positive for clarity) the displacement from the top of the cliff for the thrown rock is given by:

s = 30.0(t-2.00) +^{1}/_{2}(9.80)(t-2.00)^{2}

And the displacement for the dropped rock is simply:

s =^{1}/_{2}(9.80)t^{2}-

The ground happens to be where these two displacements are equal so:

30.0(t-2.00) +^{1}/_{2}(9.80)(t-2.00)^{2}=^{1}/_{2}(9.80)t^{2}(distribute)

30.0t-60.0 +4.9(t-2.00)^{2}= 4.9t^{2}(expand the square)

30.0t-60.0 +4.9(t^{2}-4t +4.00) = 4.9t^{2}(distribute)

30.0t-60.0 +4.9t^{2}-19.6t +19.6 = 4.9t^{2 }(Combine like terms and put all factors of t on one side)

30.0t -19.6t = 60.0-19.6^{ }10.4t = 40.4

t = 40.4/10.4 = 3.8846 s= 3.88 sb) For the height of the building let's use the displacement of the dropped stone and plug in the time of 3.8846 s

s =^{1}/_{2}(9.80)(3.8846)^{2}=73.9 mc) For the respective speeds I am going to use v = u + at where the acceleration of both is that of gravity:

for the dropped stone, u = 0:

v = u + at = 0 + 9.8(3.8846) =38.1 m/s

for the thrown stone, u = 30.0 m/s and t = 3.8846 - 2.00 = 1.8846 s:

v = u + at = 30.0 + 9.80(1.8846) =48.5 m/s

**67. A police car at rest, passed by a speeder traveling at
a constant 110 km/hr, takes off in hot pursuit. The police officer catches
up to the speeder in 700 m, maintaining a constant acceleration.
Qualitatively plot the position vs time graph for both cars from the police
car's start to the catch up point. b) calculate how long it took the police
officer to overtake the speeder, c) calculate the required police car
acceleration, and d) calculate the speed of the police car at the overtaking
point.? **

First off, 110 km/hr = 30.555 m/s

a) for the graph - here they are meeting at 700 m The speeder has a straight line plot as they are going at a constant velocity, and the police officer has a curved plot as they are accelerating uniformly:

b) The speeder's position is given by:

s = ut +^{1}/_{2}at^{2 }= ut (a = 0) = (30.555)t and since they meet at 700 m, the time is

700 = 30.555(t), t = 22.909 s =22.9 sc) The police acceleration

s = ut +^{1}/_{2}at^{2}=^{1}/_{2}at^{2 }(u = 0)

700 =^{1}/_{2}a(22.909)^{2}, a = 2.6675 m/s/s =2.67 m/s/sd) the police officer's speed is given by v = u + at:

v = 0 + (2.6675)(22.90909) = 61.111 m/s =61.1 m/s* 3.6 =220 km/hr

**69. Pelicans tuck their wings and free fall straight down
when diving for fish. Suppose a pelican starts its dive from a height of
16.0 m and cannot change its path once committed. If it takes a fish 0.20
s to perform evasive action, at what minimum height must it spot the pelican to
escape? Assume the fish is at the surface of the water. **

Before solving this wonderful problem, consider this poem by Ogden Nash:

A wonderful bird is the Pelican

His beak holds more than his belly canI know I can figure out the time it takes the pelican to reach the water: (s = -16.0m, u = 0, a = -9.80 m/s/s) using:

s = ut +^{1}/_{2}at^{2}-16.0 = 0 +^{1}/_{2}(-9.80)t^{2}t = 1.807 s

I also know that the fish needs to see the pelican .20 seconds before this or 1.807 s - .20 s = 1.607 s after the pelican has started it's descent. The pelican has gone down:

s = ut +^{1}/_{2}at^{2}= 0 +^{1}/_{2}(-9.80)(1.607)^{2}= -12.65 m and is therefore 16 - 12.65 =3.34 mabove the water.(Table of contents)

**73. Bond is standing on a bridge, 10 m above the road
below, and his pursuers are getting too close for comfort. He spots a
flatbed truck loaded with mattresses approaching at 30 m/s, which he measures by
knowing that the telephone poles are 20 m apart in this country. The bed
of the truck is 1.5 m above the road, and Bond quickly calculates how many poles
away the truck should be when he jumps down from the bridge onto the truck,
making his getaway. How many poles is it? **

Bond will fall 10 - 1.5 or 8.5 meters, and I can find the time it takes to hit the truck using (s = 8.5 m (down is +), a = 9.8, u = 0)

s = ut +^{1}/_{2}at^{2}, 8.5 = 0 +^{1}/_{2}(9.80)t^{2}, t = 1.317 s

Since the truck is going 30.0 m/s, I can use s = ut +^{1}/_{2}at^{2}(a = 0) = ut = 30.0(1.317) = 39.51 m which is almosttwotelephone poles away.(Table of contents)

Hi there!!!