Video Transcript
An object that is accelerating at three meters per second squared increases its velocity by 1.25 meters per second. For how long does the object accelerate? Give your answer to two decimal places.
Okay, so in this question we’re told that we have some object that’s accelerating at three meters per second squared. Let’s label this acceleration as 𝑎. We’re told that the velocity of the object increases by 1.25 meters per second. We’ll label the change in velocity as Δ𝑣. The fact that this value is positive shows that the velocity is increasing. We’re being asked to work out how long the object accelerates for. Let’s label this length of time over which the object accelerates as Δ𝑡. And this quantity Δ𝑡 is the thing that we’re trying to find.
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 over a time Δ𝑡, then the acceleration 𝑎 of that object is equal to Δ𝑣 divided by Δ𝑡. In this case, we know the acceleration 𝑎 of the object, and we know the change in the object’s velocity Δ𝑣. We’re trying to work out the time interval Δ𝑡 over which this velocity change occurs. So we need to take this equation for the acceleration and rearrange it to make Δ𝑡 the subject. The first step is to multiply both sides of the equation by Δ𝑡. Then on the right-hand side, we have a Δ𝑡 in the numerator which cancels with the Δ𝑡 in the denominator. This leaves us with an equation that says Δ𝑡 multiplied by 𝑎 is equal to Δ𝑣.
Then we divide both sides of the equation by the acceleration 𝑎. On the left-hand side, the 𝑎 in the numerator cancels with the 𝑎 in the denominator. So we end up with an equation that says Δ𝑡 is equal to Δ𝑣 divided by 𝑎. And the Δ𝑡 in this equation is how long the object accelerates for. In other words, it’s the quantity that the question is asking us to work out. What we need to do now is to take our values for Δ𝑣 and 𝑎 and sub them into this equation. When we do this, we get that Δ𝑡 is equal to 1.25 meters per second, that’s our value for Δ𝑣, divided by three meters per second squared, that’s our value for 𝑎. Then evaluating this expression, we find that Δ𝑡 is equal to 0.416 seconds, where the bar over the six indicates that this digit is recurring.
Looking back at the question, we see that we’re told to give our answer to two decimal places. To this precision, the result rounds up to 0.42 seconds. And so our answer to the question is that, to two decimal places, the length of time that the object accelerates for is 0.42 seconds.
