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The acceleration of a particle moving only on a horizontal

Chapter , Problem 19

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QUESTION:

The acceleration of a particle moving only on a horizontal xy plane is given by \(\vec{a}=3 t \hat{\mathrm{i}}+4 t \hat{\mathrm{j}}\), where \(\vec{a}\) is in meters per second-squared and t is in seconds. At t = 0, the position vector \(\vec{r}=(20.0 \mathrm{~m}) \hat{i}+(40.0 \mathrm{~m}) \hat{j}\) locates the particle, which then has the velocity vector \(\vec{v}=(5.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(2.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). At t = 4.00 s, what are

(a) its position vector in unit-vector notation and

(b) the angle between its direction of travel and the positive direction of the x axis?

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QUESTION:

The acceleration of a particle moving only on a horizontal xy plane is given by \(\vec{a}=3 t \hat{\mathrm{i}}+4 t \hat{\mathrm{j}}\), where \(\vec{a}\) is in meters per second-squared and t is in seconds. At t = 0, the position vector \(\vec{r}=(20.0 \mathrm{~m}) \hat{i}+(40.0 \mathrm{~m}) \hat{j}\) locates the particle, which then has the velocity vector \(\vec{v}=(5.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(2.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). At t = 4.00 s, what are

(a) its position vector in unit-vector notation and

(b) the angle between its direction of travel and the positive direction of the x axis?

ANSWER:

Step 1 of 4

Given data

Acceleration vector of the particle .

Position vector of the particle at .

Velocity vector of the particle at

Time .

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