A small ball started moving
horizontally at 16.3 meters per second. It moved in a straight line with a
uniform deceleration of three meters per square second. Determine the distance the ball
covered in the first two seconds.
The ball has a uniform
deceleration. In other words, it has a uniform
negative acceleration. That’s an indication to us that
we’re going to use the kinematic equations, the equations of uniform
acceleration. And so, the first thing that we do
is list those four equations. For an initial velocity 𝑣 naught,
an acceleration of 𝑎, a velocity of 𝑣 at time 𝑡 time units, and a displacement of
Δ𝑥, they are as shown.
Our next step is to eliminate all
but one of these equations, and we do so by listing everything we know about the
motion of our object. We’re told that the small ball
starts moving at 16.3 meters per second, so that’s its initial velocity. It has a uniform deceleration of
three meters per square second. In other words, it’s slowing
down. So, we say its acceleration is
negative three. Time 𝑡 is two seconds, and we’re
looking to find the distance that the ball covers.
We recall that distance is simply
the magnitude of the displacement. So, distance here is Δ𝑥. That’s change in 𝑥. And that means we’re not interested
in 𝑣, the final velocity of the object. So, we go through our equations and
we eliminate every equation that contains 𝑣. That’s one, three, and four,
leaving us, of course, with just equation two.
Our next job is to substitute what
we know about our ball into this equation. Δ𝑥 is what we’re trying to
calculate. Then, 𝑣 naught 𝑡 is 16.3 times
two. A half 𝑎𝑡 squared becomes a half
times negative three times two squared. Then, 16.3 times two is 32.6. And a half times negative three
times two squared is negative six. So, Δ𝑥 is 32.6 minus six, which is
26.6, and we’re measuring that in meters. So, the ball covers a distance of
26.6 meters in the first two seconds of its motion.