The initial position of an object at rest is given by 3i-8j. It moves with constant acceleration and reaches to the position 2i+4j after 4s. What is its acceleration?
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Since acceleration is constant, so we use the equation for constant acceleration
Vf = Vi + aT ⁻⁻⁻⁻⁻⁻⁻⁻⁻ (1)
Where Vi is the initial velocity
Vf the final velocity
a is acceleration
T is time taken
Given values are
Vi = 3i - 8j
Vf = 2i + 4j
T = 4 sec
a = ?
Putting these values in equation (1), we get
(2i + 4j) = (3i - 8j) + a (4)
4 a = (2i + 4j) - (3i - 8j)
4 a = (2i - 3i) + (4j + 8j)
4 a = -i + 12j
a = -1/4 i + 3 j
This is the required acceleration.
Hope it helps.
Vf = Vi + aT ⁻⁻⁻⁻⁻⁻⁻⁻⁻ (1)
Where Vi is the initial velocity
Vf the final velocity
a is acceleration
T is time taken
Given values are
Vi = 3i - 8j
Vf = 2i + 4j
T = 4 sec
a = ?
Putting these values in equation (1), we get
(2i + 4j) = (3i - 8j) + a (4)
4 a = (2i + 4j) - (3i - 8j)
4 a = (2i - 3i) + (4j + 8j)
4 a = -i + 12j
a = -1/4 i + 3 j
This is the required acceleration.
Hope it helps.
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