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1D Motion

One-dimensional Motion with Constant Acceleration

Problems for Position, Velocity, and Acceleration in One Dimension

Problems for Motion with Constant Acceleration in One Dimension

In the previous section on position, velocity, and acceleration we found that motion with constant acceleration is given by position functions of the form:

x(t) = at 2 + v 0 t + x 0

where a is the acceleration (a constant), v 0 is the velocity at time t = 0 , and x 0 is the position at time t = 0 . The velocity and acceleration functions for such a position function are given by the equations

v(t) = at + v 0 and a(t) = a.

We will now use these equations to solve some physics problems involving motion in one dimension with constant acceleration.

Free Fall

The first application we will discuss is that of objects in free fall. In general, the acceleration of an object in the earth's gravitational field is not constant. If the object is far away, it will experience a weaker gravitational force than if it is close by. Near the surface of the earth, however, the acceleration due to gravity is approximately constant--and is the same value regardless of the mass of the object (i.e., in the absence of friction from wind resistance, a feather and a grand piano fall at exactly the same rate). This is why we can use our equations for constant acceleration to describe objects in free fall near the earth's surface. The value of this acceleration is a = 9.8 m/s 2 . From now on, however, we will denote this value by g , where g is understood to be the constant 9.8 m/s 2 . (Notice that this is not valid at large distances from the surface of the earth: the moon, for instance, does not accelerate towards us at 9.8 m/s 2 .)

The equations describing an object moving perpendicular to the surface of the earth (i.e. up and down) are now easy to write. If we locate the origin of our coordinates right at the earth's surface, and denote the positive direction as that which points upwards, we find that:

x(t) = - gt 2 + v 0 t + x 0

Notice the - sign that arises because the acceleration due to gravity points downwards, while the positive position-direction was chosen to be up.

How does this relate to an object in free fall? Well, if you stand at the top of a tower with height h and let go of an object, the initial velocity of the object is v 0 = 0 , while the initial position is x 0 = h . Plugging these values into the above equation we find that the motion of an object falling freely from a height h is given by:

x(t) = - gt 2 + h

If we want to know, for instance, how long it takes for the object to reach the ground, we simply set x(t) = 0 and solve for t . We find that at t = the object hits the ground (i.e. reaches the position 0 ).

Firing a Bullet Directly Upwards

The equation

x(t) = - gt 2 + v 0 t + x 0

for an object moving up and down near the earth's surface can be used for more than just describing a falling object. We can also understand what happens to a bullet fired directly upwards from the surface of the earth at an initial speed v 0 . Since the initial position of the bullet is approximately x 0 = 0 , the equation for this motion is given by:

x(t) = - gt 2 + v 0 t

How fast will the bullet be traveling when it comes back down and hits the earth? To answer this we must (i) solve for the time at which the bullet will hit the earth, and (ii) find the velocity function, so that we can evaluate it at that time. Setting x(t) = 0 again and solving for t we find that either t = 0 or t = 2v 0/g . Well, t = 0 is just the time when the bullet left the ground, so the time at which it will come back, falling from above, must be t = 2v 0/g . Using our knowledge from the previous section, v(t) = - gt + v 0 . If we plug in t = 2v 0/g , we find that the velocity of the bullet as it comes back down and hits the ground is - g(2v 0/g) + v 0 = - v 0 . In other words, the bullet is traveling at the same speed it had when it was just fired, only in the opposite direction.

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