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Chapter 2: Problem 307

\(x\) and y co-ordinates of a particle moving in \(\mathrm{x}-\mathrm{y}\) plane at some instant are \(\mathrm{x}=2 \mathrm{t}^{2}\) and $\mathrm{y}=(3 / 2) \mathrm{t}^{2}\( Calculate y co-ordinate when its \)\mathrm{x}\( coordinate is \)8 \mathrm{~m}$. (A) \(3 \mathrm{~m}\) (B) \(6 \mathrm{~m}\) (C) \(8 \mathrm{~m}\) (D) \(9 \mathrm{~m}\)

Short Answer

Expert verified
The y-coordinate of the particle when its x-coordinate is 8 m is \(6 \mathrm{~m}\).

Step by step solution

01

Analyze the given equations for x and y

The equations given for the x and y coordinates of the particle are: \[x = 2t^2\] \[y = \frac{3}{2}t^2\]
02

Find the relationship between x and y coordinates

We can find the relationship between x and y coordinates by eliminating t from the given equations. To do that, we can express t^2 in terms of x and substitute it into the y equation. Rearranging the x equation: \[t^2 = \frac{x}{2}\] Now substitute this expression into the y equation: \[y = \frac{3}{2} (\frac{x}{2})\]
03

Simplify the relationship equation

Now, let's simplify the equation we got in Step 2: \[y = \frac{3}{2} \times \frac{x}{2}\] \[y = \frac{3}{4}x\] We have now established a relationship between the x and y coordinates of the particle.
04

Substitute x = 8 m into the equation and find the corresponding y-coordinate

Now that we have the equation relating x and y, we can substitute x = 8 m into the equation to find the corresponding y-coordinate: \[y = \frac{3}{4}(8)\] \[y = 6\] Therefore, when the x-coordinate of the particle is 8 m, its y-coordinate is 6 m. So, the correct answer is (B) \(6 \mathrm{~m}\).

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Most popular questions from this chapter

A particle moves in \(\mathrm{x}-\mathrm{y}\) plane. The position vector of the particle is given by $\mathrm{r}^{\rightarrow}=\left[3 \mathrm{ti}-2 \mathrm{t}^{2} \hat{\mathrm{j}}\right] \mathrm{m}$ Find the rate of change of \(\theta\) at \(t=1\) second. Where \(\theta\) is the angle between direction of motion and \(x\) (A) \((16 / 25)\) (B) \((12 / 25)\) (C) \(-(12 / 25)\) (D) \((16 / 9)\)

Initial angle of a projectile is \(\theta\) and its initial velocity is \(\mathrm{V}_{0}\). Find the angle of velocity with horizontal line at time \(\mathrm{t}\). (A) $\sin ^{-1}\left[1-\left\\{g /\left(\mathrm{V}_{0} \cos \theta\right)\right\\} \mathrm{t}\right]$ (B) $\tan ^{-1}\left[1-\left\\{g /\left(\mathrm{V}_{0} \cos \theta\right)\right\\} \mathrm{t}\right]$ (C) $\tan ^{-1}\left[\tan \theta-\left\\{g /\left(\mathrm{V}_{0} \cos \theta\right)\right\\} \mathrm{t}\right]$ (D) $\sin ^{-1}\left[\tan \theta-\left\\{g /\left(\mathrm{V}_{0} \cos \theta\right)\right\\} \mathrm{t}\right]$

\(\mathrm{A}^{-}+\mathrm{B}^{-}\) is perpendicular to \(\mathrm{A}^{-}\) and \(\left|\mathrm{B}^{-}\right|=2\left|\mathrm{~A}^{-}+\mathrm{B}^{-}\right|\) What is the angle between \(\mathrm{A}^{-}\) and \(\mathrm{B}^{\rightarrow}\) \((\mathrm{A})(\pi / 6)\) (B) \((5 \pi / 6)\) (C) \((2 \pi / 3)\) (D) \((\pi / 3)\)

A car goes from one end to the other end of a semicircular path of diameter ' \(\mathrm{d}\) '. Find the ratio between path length and displacement. $\begin{array}{lll}\text { (A) }(3 \pi / 2) & \text { (B) } \pi \text { (C) } 2 & \text { (D) } \pi / 2\end{array}$

\(\mathrm{t}_{1}\) and \(\mathrm{t}_{2}\) are two values of time of a projectile at the same height \(t \mathrm{t}_{1}+\mathrm{t}_{2}=\) (A) Time to reach maximum height (B) flight time for the projectile (C) \((3 / 4)\) time of the flight time. (D) \((3 / 2)\) time of the flight time.
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