Video Transcript
A particle started moving in a
straight line at seven meters per second. Given that its acceleration was of
magnitude two centimeters per second squared in the opposite direction to its
movement, find the time taken for the particle to come to rest.
In order to answer this question,
we will use the equations of constant acceleration, often known as the SUVAT
equations. The 𝑠 stands for the displacement,
the 𝑢 for the initial velocity, 𝑣 for the final velocity, 𝑎 acceleration, and 𝑡
the time. In this question, we are told that
the initial velocity is seven meters per second. The particle has acceleration of
magnitude two centimeters per second squared. However, this is in the opposite
direction to its movement. We can therefore conclude that the
particle is decelerating, and the acceleration is negative two centimeters per
second squared.
We want to calculate the time for
the particle to come to rest. This means that its final velocity
is zero. We notice that our units are
different. The velocities are given in meters
per second, whereas the acceleration is in centimeters per second squared. We recall that there are 100
centimeters in a meter. This means that we can multiply our
velocities by 100 to convert them into centimeters per second. The initial velocity is 700
centimeters per second, and the final velocity is zero centimeters per second.
The equation we will use in this
question is 𝑣 equals 𝑢 plus 𝑎𝑡 as we want the equation that doesn’t include
𝑠. Substituting in our values, we have
zero equals 700 plus negative two multiplied by 𝑡. This simplifies to 700 minus two 𝑡
is equal to zero. We can then add two 𝑡 to both
sides of this equation. Finally, by dividing both sides by
two, we get a value of 𝑡 equal to 350. The time taken for the particle to
come to rest is 350 seconds.
