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?
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Step-by-step explanation:
given :
- stone velocity = (2icap +6jcap) m/s
- ball velocity = (-icap + 2jcap) m/s.
- distance of ball and stone = 2 second
to find :
- What is the distance between the ball and the stone after 2 seconds = ?
solution :
- velocity of stone = 2î + 6j m/s
- Time of stone = 2 sec
- D = V × T
- distance of stone = (21+ 6j) × 2
- distance of stone = 4î + 12j m
- velocity of stone = -1 +2j
- Time of stone = 2 sec
then ,
D = v × t
- distance of ball = (-1 + 2j) × 2
- distance of ball = -2î + 4j m
then we have to stone and ball distance ::
stone and ball distance = (4î + 12j) - (-2i+ 4j)
stone and ball distance = 4î + 12j + 2î - 4j
stone and ball distance = (4 + 2)i + (12 - 4)j
stone and ball distance = 6î + 8j m
then, we have find stone and ball difference
stone and ball difference = √{ (6) + (8) }
stone and ball difference = √( 36 + 64)
stone and ball difference = √100
stone and ball difference = 10
vector of algebraic :
- 1. Commutative
- 2. Associative
- 3. Additive
- 4. Distributive
Answered by
1
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
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