WEBVTT
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A particle is moving in a straight line.
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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.
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Find the initial velocity of the particle π― naught and its acceleration π five seconds after it started moving.
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The displacement, π¬, is given as a vector-valued function.
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Weβre looking to find initial velocity π― naught and the acceleration of the particle five seconds after it started moving.
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So letβs begin by recalling how we link displacement, velocity, and acceleration.
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Velocity is the rate of change of displacement of an object.
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In other words, we can find an expression for the velocity by differentiating the expression for displacement with respect to time.
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This holds for vector-valued functions too.
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So the vector π― is equal to the derivative of the vector for π¬ with respect to π‘.
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Similarly, acceleration is rate of change of velocity.
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So we can differentiate our expression for velocity with respect to time to find an expression for acceleration.
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Since velocity is the first derivative of displacement with respect to time, this also means that acceleration is the second derivative of displacement.
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Thatβs d two π¬ by dπ‘ squared.
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So letβs begin by finding an expression for the velocity of the particle π― at time π‘.
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To do so, weβre going to differentiate the expression five-sixths π‘ cubed plus five π‘ with respect to π‘.
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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.
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So the derivative of five-sixths π‘ cubed with respect to π‘ is three times five-sixths π‘ squared.
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Thatβs five over two π‘ squared.
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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.
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But five π‘ to the power of zero is simply five times one, which is five.
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And so the vector π― is five over two π‘ squared plus five π.
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And then we see that its units are meters per second.
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So how can we find the initial velocity of the particle π― naught?
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Well, the initial velocity will be calculated by setting the time π‘ equal to zero.
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Thatβs five over two times zero squared plus five times the unit vector π.
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Five over two times zero squared is zero.
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So we get five π meters per second for π― naught, the initial velocity.
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Now that weβve calculated the initial velocity, letβs calculate the acceleration π five seconds after the particle starts moving.
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To do so, weβre going to begin by differentiating our expression for π― to find an expression for π at time π‘.
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The derivative of five over two π‘ squared is two times five over two π‘, which is simply five π‘.
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Then when we differentiate the constant five, we get zero.
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So our expression for acceleration π at time π‘ is five π‘ times the unit vector π.
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And its units are meters per second squared or meters per second per second.
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To find the acceleration of the particle five seconds after it starts moving, weβre going to let π‘ be equal to five.
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That gives us five times five times the unit vector π meters per square second, which is 25π meters per square second.
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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.