A particle moves along the x-axis so that at time t >= 0 its position is given by x(t)=-t^(2)+8t-9. Determine the speed of the particle at t=5. Answer:

Find Velocity Function: To find the speed of the particle at t=5, we first need to find the velocity, which is the derivative of the position function x(t) with respect to time t. The position function is x(t) = -t^2 + 8t - 9. We will calculate the derivative of x(t) to get the velocity function v(t).Calculate Derivative: The derivative of x(t) with respect to t is v(t) = dx/dt. Using the power rule, the derivative of -t^2 is -2t, the derivative of 8t is 8, and the derivative of a constant (-9) is 0. So, v(t) = -2t + 8.Evaluate at t=5: Now we need to evaluate the velocity function v(t) at t=5 to find the speed at that specific time. Substitute t=5 into v(t) to get v(5) = -2(5) + 8.Calculate Speed: Calculate v(5) by performing the arithmetic operations. v(5) = -2(5) + 8 = -10 + 8 = -2.Final Result: The velocity of the particle at t=5 is -2 cm/s. However, speed is the absolute value of velocity, as it does not have a direction. Therefore, the speed of the particle at t=5 is |v(5)| = |-2| = 2 cm/s.

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A particle moves along the
x-axis so that at time
t >= 0 its position is given by
x(t)=-t^(2)+8t-9. Determine the speed of the particle at
t=5.
Answer:

A particle moves along the $x$ -axis so that at time $t \geq 0$ its position is given by $x(t)=-t^{2}+8 t-9$ . Determine the speed of the particle at $t=5$ . $\newline$ Answer:

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Math Problems

Calculus

Relate position, velocity, speed, and acceleration using derivatives

Full solution

Q. A particle moves along the

x

-axis so that at time

t \geq 0

its position is given by

x(t)=-t^{2}+8 t-9

. Determine the speed of the particle at

t=5

\newline

Answer:

Find Velocity Function: To find the speed of the particle at $t=5$ , we first need to find the velocity, which is the derivative of the position function $x(t)$ with respect to time $t$ . The position function is $x(t) = -t^2 + 8t - 9$ . We will calculate the derivative of $x(t)$ to get the velocity function $v(t)$ .
Calculate Derivative: The derivative of $x(t)$ with respect to $t$ is $v(t) = \frac{dx}{dt}$ . $\newline$ Using the power rule, the derivative of $-t^2$ is $-2t$ , the derivative of $8t$ is $8$ , and the derivative of a constant $(-9)$ is $0$ . $\newline$ So, $v(t) = -2t + 8$ .
Evaluate at $t=5$ : Now we need to evaluate the velocity function $v(t)$ at $t=5$ to find the speed at that specific time. $\newline$ Substitute $t=5$ into $v(t)$ to get $v(5) = -2(5) + 8$ .
Calculate Speed: Calculate $v(5)$ by performing the arithmetic operations. $v(5) = -2(5) + 8 = -10 + 8 = -2$ .
Final Result: The velocity of the particle at $t=5$ is $-2$ cm/s. However, speed is the absolute value of velocity, as it does not have a direction. $\newline$ Therefore, the speed of the particle at $t=5$ is $|v(5)| = |-2| = 2$ cm/s.