Question

In space (no gravity), a stone is thrown with a velocity of (2icap +6jcap) m/s. Simultaneously, from the same point, a ball is thrown with a velocity of (-icap + 2jcap) m/s. What is the distance between the ball and the stone after 2 seconds?​

Answer 1

Answer:

• DISTANCE = 10 m

Solution:

Let the velocity at which the stone was thrown be V = 2î + 6j m/s and the velocity at which the ball was thrown be V' = -î + 2j m/s

To find the distance between the stone and the ball, first, we should find the position vector of each object after 2 seconds.

Stone:

• Velocity, V = 2î + 6j m/s
• Time, t = 2 seconds

We know,

> Displacement = Velocity × Time

> D = (2î + 6j) × 2

> D = 4î + 12j m

Stone:

• Velocity, V' = -î + 2j
• Time, t = 2 seconds

Now,

> Displacement = Velocity × Time

> D' = (-î + 2j) × 2

> D' = -2î + 4j m

Now, we have both the position vectors of each object after 2 seconds. According to the Triangle law of vector addition, If two sides of a triangle are represented by two vectors then the third side will be subtraction of the two vectors.

Hence,

> Distance b/w the stone and the ball = D - D'

> d = (4î + 12j) - (-2i + 4j)

> d = 4î + 12j + 2î - 4j

> d = (4 + 2)î + (12 - 4)j

> d = 6î + 8j m

We have the position vector of the difference between the stone and the ball, so, the magnitude of this vector will be the distance b/w the stone and the ball.

> | d | = √{ (6)² + (8)² }

> | d | = √( 36 + 64 )

> | d | = √100

> | d | = 10 m

Therefore, distance between the stone and the ball after 2 seconds will be 10 m.

Some properties of vector operation in IJ Form:

• (xî + yj + zk) × a = axî + ayj + azk

In words, when a vector is multiplied by any constant, then the resultant vector will be the same vector whose magnitude is now multiplied by that constant.

• ( xî + yj + zk ) + ( aî + bj + ck ) = (x + a)î + (y + b)j + (z + c)k

In words, the addition of two vectors will be a vector whose magnitude in each direction will be the sum of the magnitudes in respective directions of that two vectors.

Answer 2

Answer:

Given :-

In space (no gravity), a stone is thrown with a velocity of (2icap +6jcap) m/s. Simultaneously, from the same point, a ball is thrown with a velocity of (-icap + 2jcap) m/s.

To Find :-

Distance between ball and stone after 2 second

Solution :-

• For Stone

We know that

Velocity = Displacement/Time

2î + 6j = Displacement/2

2(2î + 6j) = Displacement

(2 × 2î) + (2 × 6j) = Displacement

4î + 12j = Displacement

Now

• For Ball

Velocity = Displacement/Time

-î + 2j = Displacement/2

2(-î + 2j) = Displacement

(2 × (-î)) + (2 × 2j) = Displacement

-2î + 4j = Displacement

• Finding the total displacement

4i + 12j

-2i + 4j

(+) (-)

______

6î + 8j

Finding distance between ball and stone

Distance² = (6)² + (8)²

D² = 36 + 64

D² = 100

√D² = √100

D = 10

$\rule{300}{5}$