Question Video: Rate of Velocity Change | Nagwa Question Video: Rate of Velocity Change | Nagwa

Question Video: Rate of Velocity Change Physics • First Year of Secondary School

An object increases its velocity by 2 m/s in a time of 1.25 s. What is the object’s acceleration in that time?

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Video Transcript

An object increases its velocity by two meters per second in a time of 1.25 seconds. What is the object’s acceleration in that time?

Okay, so in this question, we have an object that has some initial velocity, and we don’t know what that velocity is. But what we do know is that over a time of 1.25 seconds, that velocity increases by two meters per second. Let’s label the time interval as Δ𝑡 so that we have Δ𝑡 is equal to 1.25 seconds. We’ll also label the change in velocity as Δ𝑣 so that Δ𝑣 is equal to two meters per second. The question asks us to work out the object’s acceleration. We can recall that the acceleration of an object is defined as the rate of change of that object’s velocity.

Mathematically, if the velocity of an object changes by an amount Δ𝑣 in a time Δ𝑡, then the acceleration 𝑎 of that object is equal to Δ𝑣 divided by Δ𝑡. In this case, we know that Δ𝑣 is two meters per second and Δ𝑡 is 1.25 seconds. So we can go ahead and substitute those values into this equation to calculate the object’s acceleration. When we do this, we find that the acceleration 𝑎 is equal to two meters per second, that’s our value for Δ𝑣, divided by 1.25 seconds, that’s our value for Δ𝑡.

Evaluating this expression gives a result of 1.6 meters per second squared. Since the value of the acceleration 𝑎 is positive, this means that it’s in the same direction as the object’s velocity. And this value that we have calculated is our answer to the question. We have found that the object has an acceleration of 1.6 meters per second.

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