### Video Transcript

A particle moves along the
π₯-axis. At a time of π‘ seconds, its
acceleration is given by π is equal to 4π‘ plus six meters per second squared, when
π‘ is greater than or equal to zero. Given that at the time π‘ is equal
to two seconds its velocity is 28 meters per second, what is the particleβs initial
velocity?

The first thing that we should
notice is that our particle is moving in a straight line along the π₯-axis. And weβre given an equation for the
acceleration of this particle. We know the acceleration is the
rate of change in the velocity. What this tells us is that the
acceleration of our particle is equal to the derivative of the velocity of the
particle with respect to time. We could then integrate both sides
of this equation with respect to π‘. We see that the integral of our
derivative function with respect to π‘ would just be equal to the velocity function
up to a constant of integration π.

We also know from the question that
our function π is equal to 40 plus six. So the integral of π with respect
to π‘ is equal to the integral of 40 plus six with respect to π‘. To integrate 40, we add one to the
exponent and then divide by this new exponent, giving us 40 squared over two, which
we can simplify to two π‘ squared. We also know that the integral of
six with respect to π‘ is just 60. And then, we add our constant of
integration π. Equating these two equations gives
us that π£ plus π is equal to two π‘ squared plus 6π‘ plus π. We can then rearrange just to see
that π£ must be equal to two π‘ squared plus 6π‘ plus π minus π. Since π and π are both just
constants of integration, we can combine these into a new constant, which we will
call π one. Giving us an equation for the
velocity of our particle. The velocity of our particle is
equal to two π‘ squared plus 6π‘ plus π one.

If we go back to the question,
weβre told that when π‘ is equal to two, the velocity of the particle is 28 meters
per second. This tells us that π£ of two must
be equal to 28. So by substituting π‘ is equal to
two into our velocity function, we see that 28 must be equal to two multiplied by
two squared plus six multiplied by two plus our constant π one. We can then rearrange and solve
this to see that π one must be equal to eight. So we have the velocity of the
particle at a time π‘ must be equal to two π‘ squared plus 6π‘ plus eight. The question asks us to find the
initial velocity of the particle, which is the velocity of the particle when the
time π‘ is equal to zero. Therefore, we can substitute π‘
equals zero into our velocity function to see that the initial velocity of our
particle must be equal to two multiplied by zero squared plus six multiplied by zero
plus eight. Which we can calculate to give us
eight, which means that we have shown that the particle has an initial velocity of
eight meters per second.