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wlad13 [49]
3 years ago
7

A particle initially located at the origin has an acceleration of = 2.00ĵ m/s2 and an initial velocity of i = 8.00î m/s. Find (a

) the vector position of the particle at any time t (where t is measured in seconds), (b) the velocity of the particle at any time t, (c) the position of the particle at t = 7.00 s, and (d) the speed of the particle at t = 7.00
Physics
1 answer:
sergejj [24]3 years ago
4 0

a. The particle has position vector

\vec r(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)t\right)\,\vec\imath+\left(\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\right)\,\vec\jmath

\vec r(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath

b. Its velocity vector is equal to the derivative of its position vector:

\vec v(t)=\vec r'(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath

c. At t=7.00\,\mathrm s, the particle has position

\vec r(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(7.00\,\mathrm s)\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)^2\,\vec\jmath

\vec r(7.00\,\mathrm s)=\left(56.0\,\vec\imath+49.0\,\vec\jmath\right)\,\mathrm m

That is, it's 56.0 m to the right and 49.0 m up relative to the origin, a total distance of \|\vec r(7.00\,\mathrm s)\|=\sqrt{(56.0\,\mathrm m)^2+(49.0\,\mathrm m)^2}=74.4\,\mathrm m away from the origin in a direction of \theta=\tan^{-1}\dfrac{49.0\,\mathrm m}{56.0\,\mathrm m}=41.2^\circ relative to the positive x axis.

d. The speed of the particle at t=7.00\,\mathrm s is the magnitude of the velocity at this time:

\vec v(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)\,\vec\jmath

\vec v(7.00\,\mathrm s)=\left(8.00\,\vec\imath+14.0\,\vec\jmath\right)\dfrac{\rm m}{\rm s}

Then its speed at this time is

\|\vec v(7.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+(14.0\dfrac{\rm m}{\rm s}\right)^2}=16.1\dfrac{\rm m}{\rm s}

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