Question Video: Calculating the Acceleration Vector as the Rate Change of Velocity | Nagwa Question Video: Calculating the Acceleration Vector as the Rate Change of Velocity | Nagwa

# Question Video: Calculating the Acceleration Vector as the Rate Change of Velocity

A particle has a velocity given by 𝑣(𝑡) = (5.0𝑡𝑖 + 𝑡²𝑗 − 2.0𝑡³𝑘) m/s. What is the particle’s acceleration vector at 𝑡 = 2.0 s? What is the magnitude of the particle’s acceleration at 𝑡 = 2.0 s?

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### Video Transcript

A particle has a velocity given by 𝑣 as a function of 𝑡 equals 5.0𝑡𝑖 plus 𝑡 squared 𝑗 minus 2.0𝑡 cubed 𝑘 meters per second. What is the particle’s acceleration vector at 𝑡 equals 2.0 seconds? What is the magnitude of the particle’s acceleration at 𝑡 equals 2.0 seconds?

Since we’re asked to solve for particle acceleration at a particular time 2.0 seconds, we know that this is an instantaneous acceleration. We can write down that given time value 2.0 seconds as well as the particle’s velocity function 𝑣 of 𝑡. We’ll use this information in part one to solve for instantaneous acceleration and in part two to solve for the magnitude of that instantaneous acceleration.

We can begin solving for the instantaneous acceleration by recalling the mathematical equation that explains that term. The instantaneous acceleration an object undergoes is equal to its change in velocity divided by its change in time, specifically the time derivative of its velocity as a function of time. We can write that our acceleration as a function of time is equal to the time derivative of velocity.

When we plug in for our velocity equation and take this time derivative, we find it’s equal to 5.0𝑖 plus 2𝑡𝑗 minus 6.0𝑡 squared 𝑘 meters per second squared. This is our generalized solution for acceleration. But we want to solve for acceleration at a particular instant in time, when 𝑡 equals 2.0 seconds. To solve for it, we’ll insert that time value everywhere that 𝑡 appears in our general acceleration equation. When we calculate this value, we find it’s equal to 5.0𝑖 plus 4.0𝑗 minus 24𝑘 meters per second squared. That’s the acceleration of our object when 𝑡 equals 2.0 seconds.

Now that we know the particle’s acceleration at that time, we wanna solve for the magnitude of that acceleration. That magnitude, which tells us how much the particle’s velocity is changing at the instant in time 2.0 seconds, is equal to the square root of the acceleration’s 𝑥-component squared plus its 𝑦-component squared plus its 𝑧-component squared.

When we look at our instantaneous acceleration expression for these components, we find the 𝑥-component is 5.0, the 𝑦-component is 4.0, and the 𝑧-component is negative 24. When we enter this expression on our calculator, we find that, to three significant figures, it’s 24.8 meters per second squared. That’s the magnitude of the particle’s acceleration when 𝑡 equals 2.0 seconds.

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