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A particle starts from the origin with a velocity of (vector) v= 2i -4j +3k and a constant acceleration of (vector) a= -4i +3j -8k, where v is in m/s and a is in m/s². What is the particle's distance from the origin after 8 seconds?

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Final answer:

The particle's distance from the origin after 8 seconds can be calculated using kinematic equations, which results in a distance of approximately 284.3 meters.

Step-by-step explanation:

To calculate the particle's distance from the origin after 8 seconds, we will use the kinematic equations for motion under constant acceleration. The initial velocity vector ℓ is given by 2i - 4j + 3k and the acceleration vector a is given by -4i + 3j - 8k. After 8 seconds, the position vector ℓ(t) can be determined from:

ℓ(t) = ℓ0 + vt + ½at2

Where ℓ0 is the initial position, v is the initial velocity, a is the acceleration, and t is time. Since the particle starts at the origin, ℓ0 = 0. Plugging in the values:

ℓ(8s) = 0 + (2i - 4j + 3k)(8s) + ½(-4i + 3j - 8k)(8s)2

Simplifying gives:

ℓ(8s) = (16i - 32j + 24k) + ½(-4i + 3j - 8k)(64)

ℓ(8s) = (16i - 32j + 24k) - (128i - 96j + 256k)

ℓ(8s) = (-112i - 128j - 232k) meters

To find the distance from the origin, calculate the magnitude of the position vector:

Distance = |ℓ(8s)| = √((-112)2 + (-128)2 + (-232)2) = 284.3 meters (rounded to one decimal place)

Thus, the particle is approximately 284.3 meters from the origin after 8 seconds.

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