Problem 2

If the swimmer swims 50 meters in 20 seconds, then the average

speed is 50 meters/20 seconds = 2.5 meters/second. (If you drive

60 miles in 2 hours what is the average speed of your car?)

If the swimmer swims 50 meters in 22 seconds, then the average

speed is 50 meters/22 seconds = 2.273 meters/second.

If the swimmer swims (50+50)=100 meters in (20+22)=42 seconds,

the average speed 100 meters/42 seconds= 2.381 seconds.

One must first make the conversion from kilometers/hour to m/sec.

There are 1000 km in a meter and 3600 seconds in an hour, so

120 km/hr x 1000 / 3600 = 33.3 meters/second. If one starts with

that number the development is more transparent.

Since the airplane starts at zero speed and takes off at 33.3 m/s,

then (as I showed you in lecture on Wednesday) the average

speed is 1/2 the final speed, or 33.3/2 = 16.7 meters/second.

Since it travels 240 meters with an average speed of 16.7 m/s, the

time it takes to get up to speed is (240m)/(16.7m/s)=14.4 sec.

(How long does it take to drive 60 miles at 30 miles/hour?)

Now that you know both the final speed (33.3 m/s), and the time

it takes to reach that speed (14.4 sec), so the acceleration is just

(33.3 m/s)/(14.4 s) = 2.3 m/s/s. (If your car accelerates from

zero to 60 miles per hour in 6 seconds, the acceleration is 10 mph/s.)

If the car accelerates by 1.5m/s each sec, in 5 sec it will have

accelerated from zero speed to a final speed of (1.5 m/s/s)x(5 sec)

= 7.5m/s. Its average speed over that interval is half the final

speed, or 7.5/2 = 3.25 m/s. It travels for 5 seconds at that average

speed, which is a distance (3.25 m/s)x(5sec)=18.75 meters.

If the car then decelerates for 3 sec at 2 m/s/s, its speed will

decrease by (2 m/s/s)x(3 s) = 6 m/s, so the final speed will be

7.5 - 6 = 1.5 m/s. The average speed during the deceleration is

(7.5 + 1.5)/2 =4.5 ms, so the distance traveled is (4.5 m/s)x(3 sec)

=13.5 sec. Adding it all up:

Distance traveled = 18.75 + 13.50 = 32.35 meters

During the acceleration phase, the rocket goes from zero velocity

to (29.4 m/s/s)x(4 sec)=117.6 m/s. The average speed is half that

or 117.6/2=58.8 m/s, so in 4 sec it rises to an altitude of

58.8 x 4 = 235.2 meters.

We can get the second phase by mentally running time backwards (rewind

the VCR) and imagining that it is dropped from an additional height such that

it achieves a speed of 117.6 m/s under gravity at the point where

the rocket engine cuts out. That corresponds to a time 117.6/9.8 = 12 sec.

The average speed is the same, 58.8 m/s so the distance is

(58.8m/s)x(12 sec)= 705 meters.

Putting it all together:

total height = 235.2 + 705 = 940.8 meters

total time = 4 + 12 = 16 seconds

Problem 42

This is a little more challenging, since the rocket has a nonzero

initial speed. The expression for the height then must be solved by

the quadratic equation. The speed achieved under the acceleration in

a time t is v=50+2t. Its average speed over that time is (50+50+2t)/2,

so the distance traveled is (50+50+2t)t/2=150. The quadratic equation

gives t=2.84 seconds. The rest is like Problem 40.