ddaeeric
2020-11-12
Answered

Assume that an object moves on a trajectory according to the equations $x\left(t\right)=t+\mathrm{cos}t,y\left(t\right)=t-\mathrm{sin}t$ . Find the magnitude of the acceleration vector.

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Latisha Oneil

Answered 2020-11-13
Author has **100** answers

Step 1

An object moves on a trajectory according to the equations

$x\left(t\right)=t+\mathrm{cos}t,y\left(t\right)=t-\mathrm{sin}t$

Differentiating with respect to t, we get

${x}^{\prime}\left(t\right)=1-\mathrm{sin}t,{y}^{\prime}\left(t\right)=1-\mathrm{cos}t$

Again differentiating with respect to t, we get

$x{}^{\u2033}\left(t\right)=-\mathrm{cos}t,y{}^{\u2033}\left(t\right)=\mathrm{sin}t$

Step 2

Now, acceleration vector is

$\overrightarrow{a}\left(t\right)=x{}^{\u2033}\left(t\right)\hat{i}+y{}^{\u2033}\left(t\right)\hat{j}$

$\overrightarrow{a}\left(t\right)=-\hat{i}\mathrm{cos}t+\hat{j}\mathrm{sin}t$

Magnitude of acceleration vector is evaluated as follows.

$\left|\overrightarrow{a}\left(t\right)\right|=\sqrt{{(-\mathrm{cos}t)}^{2}+{\left(\mathrm{sin}t\right)}^{2}}=\sqrt{{\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t}=1$

Step 3

Result: Magnitude of acceleration vector is 1 unit.

An object moves on a trajectory according to the equations

Differentiating with respect to t, we get

Again differentiating with respect to t, we get

Step 2

Now, acceleration vector is

Magnitude of acceleration vector is evaluated as follows.

Step 3

Result: Magnitude of acceleration vector is 1 unit.

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