### Video Transcript

A particle is moving in a straight line. After time π‘ seconds, where π‘ is greater than or equal to zero, the bodyβs displacement relative to a fixed point is given by π¬ equals five-sixths π‘ cubed plus five π‘ π meters, where π is a fixed unit vector. Find the initial velocity of the particle π― naught and its acceleration π five seconds after it started moving.

The displacement, π¬, is given as a vector-valued function. Weβre looking to find initial velocity π― naught and the acceleration of the particle five seconds after it started moving. So letβs begin by recalling how we link displacement, velocity, and acceleration. Velocity is the rate of change of displacement of an object. In other words, we can find an expression for the velocity by differentiating the expression for displacement with respect to time. This holds for vector-valued functions too. So the vector π― is equal to the derivative of the vector for π¬ with respect to π‘.

Similarly, acceleration is rate of change of velocity. So we can differentiate our expression for velocity with respect to time to find an expression for acceleration. Since velocity is the first derivative of displacement with respect to time, this also means that acceleration is the second derivative of displacement. Thatβs d two π¬ by dπ‘ squared. So letβs begin by finding an expression for the velocity of the particle π― at time π‘. To do so, weβre going to differentiate the expression five-sixths π‘ cubed plus five π‘ with respect to π‘.

And so we begin by recalling that to differentiate a power term, we multiply the entire term by that exponent and then reduce that exponent by one. So the derivative of five-sixths π‘ cubed with respect to π‘ is three times five-sixths π‘ squared. Thatβs five over two π‘ squared. Then if we consider five π‘ as being equal to five π‘ to the power of one, the derivative is one times five π‘ to the power of zero. But five π‘ to the power of zero is simply five times one, which is five. And so the vector π― is five over two π‘ squared plus five π. And then we see that its units are meters per second.

So how can we find the initial velocity of the particle π― naught? Well, the initial velocity will be calculated by setting the time π‘ equal to zero. Thatβs five over two times zero squared plus five times the unit vector π. Five over two times zero squared is zero. So we get five π meters per second for π― naught, the initial velocity. Now that weβve calculated the initial velocity, letβs calculate the acceleration π five seconds after the particle starts moving.

To do so, weβre going to begin by differentiating our expression for π― to find an expression for π at time π‘. The derivative of five over two π‘ squared is two times five over two π‘, which is simply five π‘. Then when we differentiate the constant five, we get zero. So our expression for acceleration π at time π‘ is five π‘ times the unit vector π. And its units are meters per second squared or meters per second per second.

To find the acceleration of the particle five seconds after it starts moving, weβre going to let π‘ be equal to five. That gives us five times five times the unit vector π meters per square second, which is 25π meters per square second. And so we see that the initial velocity π― naught is five π meters per second and the acceleration at π‘ equals five is 25π meters per square second.