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.

Question
Equations
asked 2020-11-12
Assume that an object moves on a trajectory according to the equations \(\displaystyle{x}{\left({t}\right)}={t}+{\cos{{t}}},{y}{\left({t}\right)}={t}−{\sin{{t}}}\). Find the magnitude of the acceleration vector.

Answers (1)

2020-11-13
Step 1
An object moves on a trajectory according to the equations
\(\displaystyle{x}{\left({t}\right)}={t}+{\cos{{t}}},{y}{\left({t}\right)}={t}-{\sin{{t}}}\)
Differentiating with respect to t, we get
\(\displaystyle{x}'{\left({t}\right)}={1}-{\sin{{t}}},{y}'{\left({t}\right)}={1}-{\cos{{t}}}\)
Again differentiating with respect to t, we get
\(\displaystyle{x}{''}{\left({t}\right)}=-{\cos{{t}}},{y}{''}{\left({t}\right)}={\sin{{t}}}\)
Step 2
Now, acceleration vector is
\(\displaystyle\vec{{{a}}}{\left({t}\right)}={x}{''}{\left({t}\right)}\hat{{{i}}}+{y}{''}{\left({t}\right)}\hat{{{j}}}\)
\(\displaystyle\vec{{{a}}}{\left({t}\right)}=-\hat{{{i}}}{\cos{{t}}}+\hat{{{j}}}{\sin{{t}}}\)
Magnitude of acceleration vector is evaluated as follows.
\(\displaystyle{\left|{\vec{{{a}}}{\left({t}\right)}}\right|}=\sqrt{{{\left(-{\cos{{t}}}\right)}^{{2}}+{\left({\sin{{t}}}\right)}^{{2}}}}=\sqrt{{{{\cos}^{{2}}{t}}+{{\sin}^{{2}}{t}}}}={1}\)
Step 3
Result: Magnitude of acceleration vector is 1 unit.
0

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