2.11 HW 11
2.11.1 Problems listing
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2.11.2 Problem 6, section 1.2
Solve
Solution
This is separable ODE. Integrating both sides gives
Where is constant of integration. To integrate , let . Hence or . Therefore the integral becomes
But , hence the above becomes Substituting the above in (1) gives
The constant is from initial conditions. Since then Eq (2) becomes
Hence . Therefore the solution (2) becomes
2.11.3 Problem 8, section 1.2
Solve
Solution
This is separable ODE. Integrating both sides gives
The constant is from initial conditions. Since then (1) becomes
Hence the solution (1) becomes
2.11.4 Problem 24, section 1.2
A ball is dropped from the top of a building ft high. How long does it take to reach the ground?
With what speed does the ball strike the ground?
Solution
Let the ground by level (i.e. ) and let up be positive and down negative. Therefore ft and
assuming initial velocity is zero then .. Therefore Where is the acceleration, which in this case is
ft/sec. The above becomes
And
But ft. The above becomes To find the time it takes to hit the ground, the above is solved for .
This gives
Therefore the time is seconds. Now we know how long it takes to reach the ground, we
can find the velocity when ball strike the ground from (1). Substituting in (1) gives
So it strikes the ground with speed ft/sec in the downwards (negative) direction.
2.11.5 Problem 26, section 1.2
A projectile is fired straight upward with an initial velocity of m/s from the top of a building m
high and falls to the ground at the base of the building. Find (a) its maximum height
above the ground (b) when it passes the top of the building (c) its total time in the
air.
Solution
2.11.5.1 Part a
Let the ground be level . (i.e. ) and let up be positive and down negative. Therefore m. Initial
velocity is m/s, hence . The acceleration due to gravity is m/s.
When the ball reaches maximum high above the building, it must have zero velocity. From the
above this means
The above is how long it takes for the ball to reach maximum high. Now
But . Therefore Substituting in the above, gives the distance traveled above the ground until the
ball reached maximum high. Therefore
Using the above gives
2.11.5.2 Part b
The ball will take the same amount of time to fall down back to top of building, as the time it
took to reach the maximum high above the building, since the distance is the same, and the
acceleration is the same (gravity acceleration). This time is sec found in part (a). Therefore,
twice this time gives
2.11.5.3 Part c
Now we find the time it take to reach the ground. We now take initial velocity as , which is when
the ball was at its maximum high above the building. And initial position is from part (a) was
found meter. Hence m. Now we will find the time to reach the ground, starting from the
maximum high.
And
When it hits the ground ,. Hence we now have an equation to solve for time But . The above
becomes
Hence sec. This is the time it takes to fall to the ground, starting from maximum high. Adding
to this time, the time it took to reach maximum high from top of building, which is sec as found
from part (a), gives total time in air
2.11.6 Problem 5, section 1.3
Solution
Figure 2.12:Shoiwng 3 solution curves with different initial conditions
2.11.7 Problem 9, section 1.3
Solution
Figure 2.13:Shoiwng 3 solution curves with different initial conditions
2.11.8 Additional problem 1
A racecar accelerates from stationary at a rate of m/s. How long does it take the car to reach its
top speed of km/h? How far does the car travel in that time?
Solution
Let and m/s.
Since we want to find time to reach km/h which in SI units is m/sec. Substituting this in the
above gives
To find the distance traveled in this time, since
When the above gives
2.11.9 Additional problem 2
The car is approaching a tight turn at km/h. In order to safely make the corner, it must be
traveling at km/h when it enters the corner. The brakes on the car cause a deceleration
of m/s. How far away from the corner must the driver begin braking to make the
corner?
Solution
In SI units km/h is m/s. And km/h is m/s. Therefore we have initial velocity m/s and final
velocity m/s and have acceleration of m/s.
We first find the time it takes to go from to . Since
Therefore we have the equation
This is the time needed to decelerate from km/h to km/h. Now we find the distance traveled
during this time. Since
Let , by taking initial position as zero. Replacing in the above with found earlier gives
Therefore the car needs to be meter away from corner to begin the braking.
2.11.10 Additional problem 3
At the exit of the corner, two cars are traveling at km/h, with car being m behind car . Out of
the corner, car accelerates at m/s and car accelerates at m/s. How much time does it take for
car to be right next to car ? How fast are the cars going when this happens? How far from the
corner exit have they traveled?
Solution
Using SI units, km/h is m/s. Let at , and therefore , since car is ahead by meters initially. Let
m/s and also m/s. We now need to determine the time, say , where . But for car we have
And
Since . Now we do the same for car
And
Since m. Now we solve for by equating (1) and (2)
So it takes sec for car to be be next to car . To find the speed at this time, we substitute this
value of time back in the velocity equation above. For car
And for car
To find the distance traveled during this time, we substitute this time in the position equation.
For car , from Eq (1)
The distance traveled by car is meters less than this value, since it was ahead by meters at the
start at time .
2.11.11 key solution for HW11
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