A particle is moving in a straight
line such that its displacement 𝑠 after 𝑡 seconds is given by 𝑠 is equal to two
𝑡 cubed plus two 𝑡 plus two meters, where 𝑡 is greater than zero. Determine the acceleration 𝑎 as a
function of time.
In order to answer this question,
we recall that in order to find an expression for the velocity, we can differentiate
an expression in terms of time for the displacement. Likewise, we can differentiate the
velocity expression to calculate an expression for the acceleration. This means that 𝑣 of 𝑡 is equal
to d by d𝑡 of 𝑠 of 𝑡. Likewise, 𝑎 of 𝑡 is equal to d by
d𝑡 of 𝑣 of 𝑡. In this question, we are told that
the displacement 𝑠 is equal to two 𝑡 cubed plus two 𝑡 plus two meters.
Differentiating this with respect
to 𝑡 will give us an expression for the velocity. Differentiating two 𝑡 cubed gives
us six 𝑡 squared. Differentiating two 𝑡 gives us
two, and differentiating the constant two gives us zero. Therefore, 𝑣 is equal to six 𝑡
squared plus two. As the displacement is measured in
meters and the time in seconds, the velocity will be measured in meters per
second. 𝑣 is equal to six 𝑡 squared plus
two meters per second.
Differentiating this new expression
with respect to 𝑡 will give us an expression for 𝑎. When we differentiate six 𝑡
squared, we get 12𝑡, and once again differentiating a constant gives us zero. The acceleration 𝑎 is therefore
equal to 12𝑡. We can therefore conclude that if
𝑠 is equal to two 𝑡 cubed plus two 𝑡 plus two meters, then the acceleration 𝑎 of
the particle as a function of time is equal to 12𝑡 meters per second squared.