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

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A skydiver accelerates downward at a rate of 9.8 m/s². How much does their downward velocity increase in 0.67 seconds? Round your answer to two decimal places.

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### Video Transcript

A skydiver accelerates downward at a rate of 9.8 meters per second squared. How much does their downward velocity increase in 0.67 seconds? Round your answer to two decimal places.

Right, so in this question, we’ve been told that we’ve got a skydiver who’s accelerating downward. We’ve been told that their accelerating downward at a rate of 9.8 meters per second squared. So let’s say here on the side that the acceleration of the skydiver, which we’ll call 𝑎, is 9.8 meters per second squared. What we’ve been asked to do is to find out how much that downward velocity increases in 0.67 seconds. So in a time of 0.67 seconds, which we’ll call 𝑡, we need to work out the change, or more specifically, the increase in velocity of the skydiver. We’ll call this quantity Δ𝑣.

Now, it’s at this point that we can recall that the acceleration of an object, 𝑎, is defined as the change in velocity of an object divided by the amount of time taken for that velocity change to occur. Because if we think about it intuitively, what even is an acceleration? Well, an acceleration is either a speeding up or a slowing down of an object. In other words, it’s how quickly the velocity of an object increases or decreases. Also, known as the change in velocity divided by time.

Now, in this question, we’ve been given the acceleration and the amount of time for which the skydiver is accelerating. And we need to find Δ𝑣, the increase in velocity. But of course, because the acceleration is positive, we know that the velocity is going to increase. So the change in velocity which we’ll find is automatically the increase in velocity. So let’s go about finding the change in velocity.

To do this, we need to rearrange the equation by multiplying both sides by the time 𝑡. Doing this leaves us with acceleration multiplied by the time on the left-hand side and the change in velocity on the right-hand side. When we substitute in our values for the acceleration and for the time, we find that Δ𝑣 is equal to our product of the acceleration, 9.8 meters per second squared, and the time which is 0.67 seconds. Evaluating the left-hand side gives us a value of 6.566 meters per second.

However, this is not our final answer. Remember, the last sentence of the question tells us to round our answer to two decimal places. So here’s, the first decimal place, that five. And here’s the second decimal place which is a six. We need to work out what to do with that six. To work this out, we need to look at the next decimal place which happens to be also six in this case. Now, this third decimal place is six, as we said, and six is larger than five. Therefore, the second decimal place will round up. It’s gonna turn into a seven. And so, to two decimal places, we have 6.57 meters per second.

At which point, we found our final answer. The downward velocity of a skydiver increases by 6.57 meters per second, to two decimal places.

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