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Wed 17 Apr
Applications of integration: Unit test
A particle with velocity
v(t)=3sqrtt, where
t is time in seconds, moves in a straight line.
How far does the particle move from
t=1 to
t=4 seconds?

◻ units

Wed $17$ Apr $\newline$ Applications of integration: Unit test $\newline$ A particle with velocity $v(t)=3 \sqrt{t}$ , where $t$ is time in seconds, moves in a straight line. $\newline$ How far does the particle move from $t=1$ to $t=4$ seconds? $\newline$ $\square$ units

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Relate position, velocity, speed, and acceleration using derivatives

Full solution

Q. Wed

17

Apr

\newline

Applications of integration: Unit test

\newline

A particle with velocity

v(t)=3 \sqrt{t}

, where

t

is time in seconds, moves in a straight line.

\newline

How far does the particle move from

t=1

to

t=4

seconds?

\newline

\square

units

Integrate velocity function: To find the distance the particle moves, we need to integrate the velocity function from $t=1$ to $t=4$ .
Set up integral: Set up the integral of the velocity function $v(t) = 3\sqrt{t}$ from $t=1$ to $t=4$ .
Calculate integral: Calculate the integral: $\int_{1}^{4} 3\sqrt{t} \, dt$ .
Find antiderivative: The antiderivative of $3\sqrt{t}$ is $3 \times \left(\frac{2}{3}\right)t^{\frac{3}{2}}$ , which simplifies to $2t^{\frac{3}{2}}$ .
Evaluate antiderivative: Evaluate the antiderivative from $t=1$ to $t=4$ : $[2t^{(3/2)}]$ from $1$ to $4$ .
Plug in upper limit: Plug in the upper limit: $2(4)^{\frac{3}{2}} = 2(8) = 16$ .
Plug in lower limit: Plug in the lower limit: $2(1)^{\frac{3}{2}} = 2(1) = 2$ .
Subtract limits: Subtract the lower limit from the upper limit: $16 - 2 = 14$ .

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