A particle leaves the origin with initial velocity v→=3i^ m s^(-1) and a constant acceleration a→=-1i^-0.5j^m s^(-2). The position vector of particle, when it reaches its maximum x-coordinate, is

Correct answer: (B) 9/2i^-(j^)/2 m. Using the first equation of motion, v=u+at, in-plane, v→t=3-1×ti^+0-0.5tj^ ...i for maximum positive, x coordinate when v_x become zero. ∴ 3-t=0⇒t=3 s than r→3=4.5i^-2.25j^=9/2i^-(j^)/2 m

Q126 / 160 · TS EAMCET 2021 (05 Aug Shift 1) · Physics

A particle leaves the origin with initial velocity $\vec{v} = (3 \hat{i}) m s^{- 1}$ and a constant acceleration $\vec{a} = (- 1 \hat{i} - 0.5 \hat{j}) m s^{- 2}$ . The position vector of particle, when it reaches its maximum $x$ -coordinate, is

A
$\frac{9}{2} (\hat{i} - \hat{j}) m$
Correct answer
B
$\frac{9}{2} (\hat{i} - \frac{\hat{j}}{2}) m$
C
$\frac{9}{2} (- \hat{i} + \hat{j}) m$
D
$\frac{9}{2} (\frac{\hat{i}}{2} - \hat{j}) m$

Solution

Using the first equation of motion, $v = u + a t$ , in-plane,

$\vec{v} (t) = (3 - 1 \times t) \hat{i} + (0 - 0.5 t) \hat{j} . . . (i)$

for maximum positive, $x$ coordinate when $v_{x}$ become zero.

$∴ 3 - t = 0 \Rightarrow t = 3 s$

than $\vec{r} (3) = 4.5 \hat{i} - 2.25 \hat{j} = \frac{9}{2} (\hat{i} - \frac{\hat{j}}{2}) m$

Hence, the correct answer is option: 'B'.

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A particle leaves the origin with initial velocity v→=3i^ m s-1 and a constant acceleration a→=-1i^-0.5j^m s-2. The position vector of particle, when it reaches its maximum x-coordinate, is

Solution

Relevant Questions

A particle leaves the origin with initial velocity $\vec{v} = (3 \hat{i}) m s^{- 1}$ and a constant acceleration $\vec{a} = (- 1 \hat{i} - 0.5 \hat{j}) m s^{- 2}$ . The position vector of particle, when it reaches its maximum $x$ -coordinate, is