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.

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.