Q: "A driver going from Salt Lake City to Provo, Utah, a distance of 60 miles, wants to maintain an average speed of 65 miles/hour. The freeway in Salt Lake is under construction and so for the first 30 miles the driver averages only 35 miles/hour. How fast does he need to go remaining 30 miles in order to meet his 65 mile/hour goal?" A: The driver's average speed between Salt Lake and Provo is determined by taking the sum of all the various speeds the driver experiences and dividing it by the number of speeds. More clearly, in mathematical terms,: v_{total} = (v_{1} + v_{2} + v_{3} + ... + v_{n} ) / n In our case, n = 2 since there are only two different speeds. The equation becomes: v_{total} = (v_{1} + v_{2}) / 2 We know the total speed, v_{total} = 65 mi/hr, and the first velocity, v_{1} = 35 mi/hr. We need to find the second velocity, v_{2}, that will make the final velocity equal to 65 mi/hr. Rearranging the equation above: v_{total} = (v_{1} + v_{2}) / 2 or v_{2} = (2 v_{total} )  v_{1} Solve for v_{2}: v_{2} = (2 x 65 mi/hr)  (35 mi/hr) = (130  35) mi/hr = 95 mi/hr In order for the driver to make it
to Provo with an average speed of 65 mi/hr, he will need to travel 95
mi/hr for the last 30 miles of his trip. Better watch for speed
traps. 

Q: "A hotdog pilot flies his airplane 10 meters above the ground. He encounters a hill that slopes upward at 3.4 degrees. If the pilot has just 1.5 seconds to correct his path before he hits the hill, how fast is he going? A: The first step is to draw a picture: To find the horizontal distance, d, from the bottom of the hill to the point where the hill intersects the original flight path of the plane, we look at the triangle defined by the flight path, the face of the hill, and the height, h, that the plane is flying. Now the tangent function can be used to determine the distance. tan 3.4º = h / d or d = h / tan 3.4º Solve for d: d = 10 m / tan 3.4º = 168.32 m. The plane will crash into the hill 168.32 meters after it passes over the bottom of the hill. We know the distance from the bottom of the hill to the point where the plane will crash (d = 168.32 m) and the time it takes to cover this distance (t = 1.5 sec). Using the equation d = v t, we can find the plane's velocity d = v t or v = d / t v = 168.32 m / 1.5 sec = 112.2 m / sec The plane is traveling at a
velocity of 112 m/sec in the horizontal direction before it makes the
course correction. 

Q: "Can an object be increasing in speed as its acceleration decreases?" A:
I assume you mean the acceleration is positive, so that the velocity
increases. The answer to your question is then YES. As long
as the acceleration remains positive, even if it is not constant (it is
increasing or decreasing), the velocity will keep increasing. 

Q: "A ball is thrown straight upward from the top of a cliff with an initial velocity of 15 m/sec. Assuming the ball just misses the edge of the cliff on its way down, how fast is it going just before it hits the ground which is 35 m below the ball's starting point?"
To find out how far the ball has traveled in the first 1.5 sec we use the equation: x = x_{0} + v_{0} t + (1/2) at^{2}.
Part II: The maximum height of the ball is 46.5 m (35 m + 11.5 m) above the bottom of the cliff. Since the velocity (v) of the ball at this point is 0 m/sec, we can treat the ball as if it had been dropped from rest at 46.5 m above the cliff bottom. To find the velocity of the ball just before it hits the ground, we need to find the time it takes the ball to hit the ground. Again we use the equation x = x_{0} + v_{0} t + (1/2) at^{2}.
Finally, to determine the velocity of the ball dropped from 46.5 m and falling for 3.1 sec with the acceleration of 9.8 m/sec^{2}, we can again use the equation v = v_{0} + at.


Q: "If I run 400 meters in 30 seconds, what is my average velocity?" A: The
average velocity (v) is equal to the change in position (x = x_{2}
 x_{1}) divided by the change in time (t
= t_{2}  t_{1}). = (400 m  0 m) / ( 30
sec  0 sec) = 400 m / 30 sec = 13.33 m/s 

Q: Can an object have zero velocity and still be accelerating? A: Yes,
acceleration is the change in velocity over time. If an object has
zero velocity at one instant (t) but changes is velocity
in the next, then the object is still accelerating. Similarly if
an object is changing direction but it's velocity is 0 or constant it is
still accelerating. Only when the object stops changing direction
and speed (Newton's 1st Law) does acceleration = 0. 

Q: Can an object have a constant velocity and still have a varying speed? A: No,
speed is velocity without direction (the magnitude of velocity).
If the velocity is constant than the speed is also constant.
Similarly the speed is constant if the object is only undergoing
directional changes. 

Q: Can the velocity of an object reverse direction when the object's acceleration is constant? A:
Yes, acceleration is the rate at which the velocity is changing.
If the object's velocity was switching from positive to negative and
back again then the object would have a constant acceleration. 
[Top] [Previously Asked Questions] [References]
Displacement  x = x_{2}  x_{1} 
Average Velocity  
Average Speed  
Instantaneous Velocity  
Average Acceleration  
Instantaneous Acceleration  
Kinematic
Equations for Motion with Constant Acceleration
(These are not valid when acceleration is not constant) 
v = v_{0}
+ at
x  x_{0} = v_{0} t + at^{2} x  x_{0} = (v_{0} + v)t x  x_{0} = vt  at^{2} 
[Top] [Previously Asked Questions] [References]
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