Question Video: Component Form of the Acceleration of a Particle

A particle accelerates uniformly. At the time 𝑑 = 0.0 s, the particle has a velocity 𝑣 = (14𝑖 + 22𝑗) m/s. At 𝑑 = 3.8 s, the particle has a velocity 𝑣 = (0.0𝑖 + 11𝑗) m/s. What is the acceleration of the particle?

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Video Transcript

A particle accelerates uniformly. At the time 𝑑 equals 0.0 seconds, the particle has a velocity 𝑣 equals 14𝑖 plus 22𝑗 meters per second. At 𝑑 equals 3.8 seconds, the particle has a velocity 𝑣 equals 0.0𝑖 plus 11𝑗 meters per second. What is the acceleration of the particle?

Since we’re solving for the acceleration of the particle over a time interval from 𝑑 equals 0.0 to 3.8 seconds, we know that we’re solving for an average acceleration. To begin on our solution, we can recall the mathematical relationship for average acceleration. The average acceleration of an object is equal to its final velocity minus its initial velocity divided by the time interval over which that velocity change happens.

In our case, we could write that our initial time 𝑑 sub 𝑖 is 0.0 seconds. Our initial velocity is 14𝑖 plus 22𝑗 meters per second. And our final time is 3.8 seconds, and our final velocity is 0.0𝑖 plus 11𝑗 meters per second. If we calculate the difference 𝑣 sub 𝑓 minus 𝑣 sub 𝑖, we find, as we treat these vectors separately by their 𝑖 and 𝑗 components, that we end up with a total vector of negative 14𝑖 minus 11𝑗 meters per second.

So, when we go to calculate our average acceleration, we have this vector as our numerator divided by our time difference of 𝑑 sub 𝑓, 3.8 seconds, minus 𝑑 sub 𝑖, 0.0 seconds. Our total time interval then is 3.8 seconds. And when we calculate this fraction, we find a result of negative 3.7𝑖 minus 2.9𝑗 meters per second squared. That’s the acceleration experienced by the particle over this time interval.

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