The position of a particle is given by r=2.00t^3i + (t^2-1.00)j where [t]=s and [r]=m. Find the magnitude of the velocity of this particle at time t=1.20s.

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The position of a particle is given by r=2.00t^3i + (t^2-1.00)j where [t]=s and [r]=m. Find the magnitude of the velocity of this particle at time t=1.20s.

Physics
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plug in 1.2s to get r in m. then since you know velocity=m/s divide the distance travelled by time. or you can try taking the derivative to get velocity and then plug in your time to directly get it :)
I did it by taking the derivative and plugging the time in....I got 8.75m/s but I then did i the first way and got 2.90m/s....which is right?
i think it is the second way sry the first way cant work like that. i only looked at the numbers to come up with a solution i read it. it should b the second way :)

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Okay thanks :)
I got about 8.97... \[v(t)=6.00t^2i+2tj\]\[v(1.2)=6(1.2)^2i+2(1.2)j=8.64i+2.4j\]\[|v|=\sqrt{8.64^2+2.4^2}\approx8.97\]
Your initial intuition is correct. Take the derivative of r w.r.t. time. Plug in time. Get velocity.

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