Motion of Point-Mass Objects in One Dimension

archerA.gif (32597 bytes)

Previously Asked Questions

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,:

vtotal = (v1 + v2 + v3 + ... + vn ) / n

In our case, n = 2 since there are only two different speeds.  The equation becomes:

vtotal = (v1 + v2) / 2

We know the total speed, vtotal = 65 mi/hr, and the first velocity, v1 = 35 mi/hr.  We need to find the second velocity, v2, that will make the final velocity equal to 65 mi/hr.  Rearranging the equation above:

vtotal = (v1 + v2) / 2     or    v2   =  (2 vtotal ) - v1

Solve for v2:

v2 = (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:

plane.gif (4564 bytes)

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?"

cliff.gif (26670 bytes) A:    To solve this problem we first draw a diagram of the problem.

Then, we will divide the problem into two parts.  In the first part the ball is traveling upward.  In the second part the ball is falling to the bottom of the cliff.

Part I:

When the ball reaches its maximum height its velocity (v) must be 0, since it is traveling neither up nor down at this point.  To determine the time (t) it takes the ball to reach its highest point, we can use the equation v = v0 + at.

= v0 + at  ->  ( v - v0 ) / a =   t Derive an equation for t from v = v0 + at.
v0 = 15 m/sec The initial velocity given in the problem
= - 9.8 m/sec2 The acceleration of gravity (g).
v = 0 m/sec The velocity at the maximum height
t = ( v - v0 ) / a Solve for t
= ( 0 m/sec - 15.0 m/sec) / ( - 9.8 m/sec2)
= (15 m/sec) / (9.8 m/sec2)
= 1.5 sec.
So, after 1.5 sec the ball stops traveling upward.

To find out how far the ball has traveled in the first 1.5 sec we use the equation: x = x0 + v0 t + (1/2) at2.

x0 = 0 m. The initial position of the ball which, for simplicity, we will take as 0 m.
v0 = 15 m/sec The initial velocity given in the problem
t   = 1.5 sec Determined above
= - 9.8 m/sec2 The acceleration of gravity (g).
x = x0 + v0 t + (1/2) at2 Solve for x
= 0 m + (15 m/sec) (1.5 sec) + (1/2)(- 9.8 m/sec2)(1.5 sec)2
= 22.5 m. + (0.5)(-9.8 m/sec2)(2.3 sec2)
= 22.5 m - (11.0) m
= 11.5 m
From the initial point the ball travels 11.5 m upward before falling to the bottom of the cliff.

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 = x0 + v0 t + (1/2) at2.

x0 = 46.5 m. The ball is being dropped from a height of 46.5 m
v0 = 0 m/sec The initial velocity of the ball at its maximum height is 0 m/sec
a = - 9.8 m/sec2 The acceleration of gravity (g).
= 0 m. The balls final position is 0 m from the ground.
x = x0 + v0 t + (1/2) at2  Derive an equation for t from
x = x0 + v0 t + (1/2) at2.
0 m = x0 + (0 m/sec) t + (1/2) at2 Use the facts that v0 = 0 m/sec and
x = 0 m to simplify the derivation.
0 m = x0 + (1/2) at2
-> 2 (-x0) / at2
- > t = sqrt.gif (76 bytes)( - 2 x0 / a )
t = sqrt.gif (76 bytes)( - 2 x0 / a ) Solve for t
= sqrt.gif (76 bytes)( - 2 (46.5 m) / (- 9.8 m/sec2)
= sqrt.gif (76 bytes)( 9.478 sec2)
= 3.1 sec
The ball takes 3.1 sec to fall 46.5 m.

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/sec2, we can again use the equation v = v0 + at.

v0 = 0 m/sec The ball has no velocity at its highest point.
a = - 9.8 m/sec2 The acceleration of gravity (g).
t   = 3.1 sec Determined above
v = v0 + at Solve for v
= (0 m/sec) + (- 9.8 m/sec2) (3.1 sec)
= - 30.2 m/sec The negative sign indicates the direction of velocity is downward.
The ball has a velocity of 30.2 m/sec downward just before it hits the ground.

Note: The mass of the ball did not matter in this problem because all objects on Earth fall at the same rate (g).

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 (delta.gif (839 bytes)x = x2 - x1) divided by the change in time (delta.gif (839 bytes)tt2 - t1). 
In this case we take the initial position x1 =  0 meters and the final position x2 = 400 meters.
For time we let the initial time t1 = 0 seconds and the final time t2 = 30 seconds.  Then we solve the equation:

2-2.gif (1084 bytes) =  (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.

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Displacement delta.gif (839 bytes)x = x2 - x1
Average Velocity 2-2.gif (1084 bytes)
Average Speed 2-3.gif (1107 bytes)
Instantaneous Velocity 2-4.gif (1152 bytes)
Average Acceleration

2-7.gif (961 bytes)

Instantaneous Acceleration 2-8.gif (957 bytes)

2-9.gif (990 bytes)

Kinematic Equations for Motion with Constant Acceleration

(These are not valid when acceleration is not constant)

v = v0 + at

x -  x0 = v0 t + onehalf.gif (67 bytes)at2

2-11.gif (938 bytes)

x -  x0 = onehalf.gif (67 bytes)(v0  + v)t

x -  x0 = vt - onehalf.gif (67 bytes) at2

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Mechanics List of Topics

Measurements Newton's Laws Potential Energy and Conservation of Energy Rotation of
Rigid Bodies
Vectors Forces and Fields Linear Momentum Angular Momentum Mechanical
Motion of Point-Mass Objects in One Dimension The Gravitational Field Collisions Torque Mechanical Waves
Motion of Point-Mass Objects in Two and Three Dimensions Kinetic Energy
and Work
Circular Motion of Point-Mass Objects Equilibrium Sound