The position of a particle moving in an \(x y\) plane is given by \(\overrightarrow{\mathbf{r}}=\left[\left(2 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3}-(5 \mathrm{~m} / \mathrm{s}) t\right] \hat{\mathbf{i}}+\left[(6 \mathrm{~m})-\left(7 \mathrm{~m} / \mathrm{s}^{4}\right) t^{4}\right] \hat{\mathbf{j}} . \quad\) Cal- culate \((a) \overrightarrow{\mathbf{r}},(b) \overrightarrow{\mathbf{v}}\), and \((c) \overrightarrow{\mathbf{a}}\) when \(t=2 \mathrm{~s}\).

Short Answer

Expert verified
After taking the steps, we obtain the values for: \n(a) \(\overrightarrow{\mathbf{r}}\) at t=2s \n(b) \(\overrightarrow{\mathbf{v}}\) at t=2s \n(c) \(\overrightarrow{\mathbf{a}}\) at t=2s.

Step by step solution

01

Find Position Vector at t=2s

To find the position vector at \( t=2 s \), we substitute \( t=2 \) into the position vector. \n \[ \overrightarrow{\mathbf{r}}=\left[\left(2 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3}-(5 \mathrm{~m} / \mathrm{s}) t\right] \hat{\mathbf{i}}+\left[(6 \mathrm{~m})-\left(7 \mathrm{~m} / \mathrm{s}^{4}\right) t^{4}\right] \hat{\mathbf{j}} \]
02

Differentiate position vector to find velocity

Differentiate the position vector with respect to time to find the velocity vector. Remember to apply the power rule for differentiation, d/dx[x^n] = nx^{n-1}. Apply this rule to each term in the position vector to find the velocity vector.
03

Evaluate velocity vector at t=2s

Substitute \( t = 2 s \) into the equation for the velocity vector to find the velocity of the particle at \( t = 2 s \).
04

Differentiate velocity vector to find acceleration

Differentiate the velocity vector with respect to time to find the acceleration vector. Use the chain rule for differentiation, d/dx[f(g(x))] = f'(g(x)) * g'(x). Apply this rule to each term in the velocity vector to get the acceleration vector.
05

Evaluate acceleration vector at t=2s

Substitute \( t = 2 s \) into the equation for the acceleration vector to find the acceleration of the particle at \( t = 2 s \).

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