Video Transcript
A particle moves in a straight line
with an initial velocity of 30.0 meters per second. It accelerates for 5.00 seconds at
a constant 30.0 meters per second squared in the direction of its initial
velocity. What is the magnitude of the
particle’s displacement after the acceleration? What is the magnitude of the
particle’s velocity after the acceleration?
In this two-part problem, we want
to solve for the particle’s displacement after it undergoes its acceleration, which
we can call 𝑑. And we also wanna solve for its
velocity after this acceleration occurs. We can call this 𝑣 sub 𝑓. In terms of the particle’s motion,
we’re told it starts out with an initial velocity of 30.0 meters per second. And then, for 5.00 seconds, it
accelerates at 30.0 meters per second squared. We can come to a better
understanding of the particle’s motion by drawing a graph of its velocity versus
time. Here we have a chart of the
particle’s velocity versus time, where time, in units of seconds, is on the
horizontal axis and velocity, in units of meters per second, is on the vertical
axis.
We’re told that our particle begins
with a velocity, we’ve called it 𝑣 sub 𝑖, of 30.0 meters per second. So we mark that value in as our
initial velocity on our chart. We’re told further that our
particle that accelerates at an acceleration of 30.0 meters per second squared for a
time, we’ve called 𝑡 sub 𝑎, of 5.00 seconds. When we write out our acceleration
of 30.0 meters per second squared, we understand that that means 30.0 meters per
second every second. In other words, for every one
second of time passed, we add 30.0 meters per second to our velocity. So if the next tick mark on our
vertical axis is 60 meters per second, then we achieve that velocity at a time of
one second. And under this continuing
acceleration of 30.0 meters per second, we continue to add 30.0 meters per second to
our velocity every second.
By connecting these dots, we now
have our velocity versus time curve for the particle. And we see that under these
conditions, the maximum velocity we achieve is 180 meters per second. That’s our result for 𝑣 sub 𝑓,
the final velocity achieved. In our case, we’ve solved for this
result graphically. That’s another way of finding the
same result. Now, we want to solve for the total
displacement of the particle. We’ve called it 𝑑. We can recall that an object’s
displacement is equal to the integral of its velocity with respect to time. And our graph is a velocity versus
time curve. This means that if we solve for the
area under our velocity versus time curve, then that total area result will give us
our displacement 𝑑.
To make calculating this total area
easier, we can divide the area up into two pieces, a rectangular piece as well as a
triangular piece above that. We can now write that our
displacement 𝑑 is equal to the area of the rectangle plus the area of the triangle
in units of meters. We can write that the area of our
rectangle is equal to its height, 30.0 meters per second, times it’s width, 5.00
seconds. Notice the units of seconds cancel,
and we’re left with units of meters. Then we add that area to the area
of our triangle which is one-half its base, 5.00 seconds, times its height, 180
minus 30, or one 150 meters per second. Again, the units of seconds cancels
out, and we’re left with units of meters. When we add these two values
together, we find a result of 525 meters. That’s the particle’s total
displacement after these five seconds of acceleration and starting at our initial
velocity 𝑣 sub 𝑖.
