# Assume that an object moves on a trajectory according to the equations x(t)=t+cost, y(t)=t−sint. Find the magnitude of the acceleration vector.

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
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
$x{}^{″}\left(t\right)=-\mathrm{cos}t,y{}^{″}\left(t\right)=\mathrm{sin}t$
Step 2
Now, acceleration vector is
$\stackrel{\to }{a}\left(t\right)=x{}^{″}\left(t\right)\stackrel{^}{i}+y{}^{″}\left(t\right)\stackrel{^}{j}$
$\stackrel{\to }{a}\left(t\right)=-\stackrel{^}{i}\mathrm{cos}t+\stackrel{^}{j}\mathrm{sin}t$
Magnitude of acceleration vector is evaluated as follows.
$|\stackrel{\to }{a}\left(t\right)|=\sqrt{{\left(-\mathrm{cos}t\right)}^{2}+{\left(\mathrm{sin}t\right)}^{2}}=\sqrt{{\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t}=1$
Step 3
Result: Magnitude of acceleration vector is 1 unit.