A particle undergoes a displacement r-2-0-3-0-6-0k- ending with the position vector r-3-0-4-0k- in meters- What was its initial position vector-

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Standard XII

Physics

Displacement

A particle un...

Question

A particle undergoes a displacement $△ \to r = 2.0^i - 3.0^j + 6.0^k$ , ending with the position vector $\to r = 3.0^j - 4.0^k$ , in meters, What was its initial position vector?

$3^j - 4^k$

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$2^j - 2^k$

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$2^i - 6^j + 10^k$

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$- 2^i + 6^j - 10^k$

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Solution

The correct option is D
$- 2^i + 6^j - 10^k$

$△ r = r_{f} - r_{i}$
$r_{i} = r_{f} - △ r$
$= 3^j - 4^k - (2^i - 3^j + 6^k)$
$r_{i} = - 2^i + 6^j - 10^k$

So the particle started from the initial position whose position vector is given by
$r_{i} = - 2^i + 6^j - 10^k$

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Similar questions

A particle undergoes a displacement $△ \to r = 2.0^i - 3.0^j + 6.0^k$ , ending with the position vector $\to r = 3.0^j - 4.0^k$ , in meters, What was its initial position vector?

What are the initial position vector ${\to r}_{i}$ and final position vector ${\to r}_{f}$ , both in unit-vector notation? What is the x component of displacement $Δ \to r$ ?

(i)	Position vector ${\to r}_{i}$	(x)	$5^i - 3^j - 1^k$
(ii)	Postion vector ${\to r}_{f}$	(y)	$- 2^i - 4^j + 1^k$
(iii)	x-component of displacement $Δ \to r$	(z)	$7^i + 1^j - 2^k$

Q. In a two dimensional motion, a particle moves from point A (position vector

{\to r}_{1}

), to point B (position vector

{\to r}_{2}

). If the magnitudes of these vectors are

r_{1} = 3 m

and

r_{2} = 4 m

and the angles they make with the x-axis are

θ_{1} = 75^{\circ} and θ_{2} = 15^{\circ}

, respectively, then the magnitude of the displacement vector (in

m

) is

In a two dimensional motion of particle, the particle moves from point Α, position vector ${\to r}_{1}$ to point B, position vector ${\to r}_{2}$ . If the magnitudes of these vectors are, respectively $r_{1} = 3 a n d r_{2} = 4$ , and the angles they make with the x- axis are $θ_{1} = 75^{0} a n d θ_{2} = 15^{0},$ and respectively, then find the magnitude of the displacement vector.

What are the initial position vector ${\to r}_{i}$ and final position vector ${\to r}_{f}$ , both in unit-vector notation? What is the x component of displacement $Δ \to r$ ?

(i)	Position vector ${\to r}_{i}$	(x)	$5^i - 3^j - 1^k$
(ii)	Postion vector ${\to r}_{f}$	(y)	$- 2^i - 4^j + 1^k$
(iii)	x-component of displacement $Δ \to r$	(z)	$7^i + 1^j - 2^k$

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A particle undergoes a displacement △→r=2.0^i−3.0^j+6.0^k, ending with the position vector →r=3.0^j−4.0^k, in meters, What was its initial position vector?

The correct option is D −2^i+6^j−10^k △r=rf−ri ri=rf−△r =3^j−4^k−(2^i−3^j+6^k) ri=−2^i+6^j−10^k So the particle started from the initial position whose position vector is given by ri=−2^i+6^j−10^k

A particle undergoes a displacement $△ \to r = 2.0^i - 3.0^j + 6.0^k$ , ending with the position vector $\to r = 3.0^j - 4.0^k$ , in meters, What was its initial position vector?

The correct option is D
$- 2^i + 6^j - 10^k$

$△ r = r_{f} - r_{i}$
$r_{i} = r_{f} - △ r$
$= 3^j - 4^k - (2^i - 3^j + 6^k)$
$r_{i} = - 2^i + 6^j - 10^k$

So the particle started from the initial position whose position vector is given by
$r_{i} = - 2^i + 6^j - 10^k$