The position vector of an object moving in X-Z plane is r-v0t-a0eb0tk-Find its i velocity - v-dr-dt -ii speed - v- -iii acceleration - dv-dt - as a function of time-

The correct option is B (i) v⃗=v_0î+a_0b_0e^b_0tk̂ #10;(ii) v⃗=√(v_0^2+a_0^2b_0^2e^2b_0t) #10; (iii) a⃗=a_0b_0^2e^b_0tk̂ Velocity of an obje ...

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Condition for Coplanarity of Four Points

The position ...

Question

The position vector of an object moving in $X - Z$ plane is $\to r = v_{0} t^i + a_{0} e^{b_{0} t}^k$ .
Find its (i) velocity $(\begin{matrix} \to v = \frac{d \to r}{d t} \end{matrix})$
(ii) speed $(\begin{matrix} | \begin{matrix} \to v \end{matrix} | \end{matrix})$
(iii) acceleration $(\begin{matrix} \frac{d \to v}{d t} \end{matrix})$ as a function of time.

(i)

\to v = v_{0}^i + a_{0} b_{0} e^{b_{0} t}^k

(ii)

| \begin{matrix} \to v \end{matrix} | = \sqrt{2 v_{0}^{2} + a_{0}^{2} b_{0}^{2} e^{2 b_{0} t}}

(iii)

\to a = a_{0} b_{0}^{2} e^{b_{0} t}^k

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(i)

\to v = v_{0}^i + a_{0} b_{0} e^{b_{0} t}^k

(ii)

| \begin{matrix} \to v \end{matrix} | = \sqrt{v_{0}^{2} + a_{0}^{2} b_{0}^{2} e^{2 b_{0} t}}

(iii)

\to a = a_{0} b_{0}^{2} e^{b_{0} t}^k

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(i)

\to v = v_{0}^i + a_{0} b_{0} e^{b_{0} t}^k

(ii)

| \begin{matrix} \to v \end{matrix} | = \sqrt{v_{0}^{2} + a_{0}^{2} b_{0}^{2} e^{2 b_{0} t}}

(iii)

\to a = 2 a_{0} b_{0}^{2} e^{b_{0} t}^k

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None of the above

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