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

Question Video: Rate of Velocity Change Physics

A train has a velocity of 35 m/s when it reduces its engine power, and after 5 seconds the train has a velocity of 11.5 m/s. What is the train’s average acceleration in its direction of motion?

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Video Transcript

A train has a velocity of 35 metres per second when it reduces its engine power. And after five seconds, the train has a velocity of 11.5 metres per second. What is the train’s average acceleration in its direction of motion?

Let’s begin by underlining the important bits of the question. So we know that the train that we’re looking at was travelling initially at 35 metres per second. Then it reduces its engine power. And after five seconds, the train has a velocity of 11.5 metres per second. What we’re asked to do is to find the train’s average acceleration in the direction of motion.

So what we have is a train that initially is travelling at 35 metres per second. Then five seconds later, it’s got a speed of 11.5 metres per second. What we want to work out is its average acceleration, that is to say, how much speed it’s losing every second.

After all, acceleration is just a rate of change of speed or how much speed changes by for a given unit of time. In this case, the unit of time is one second. So we want to find out how much the speed changes by on average every second.

Essentially, the train starts out at 35 metres per second. And then after one second, it loses some speed. After two seconds, it loses some more speed, so on for the third, fourth, and fifth second. When the question asks us to find the average acceleration of the train, all that’s telling us is to assume that the change in speed every second is exactly the same.

So the first thing that we can do is to work out the total change in speed of the train when it reduces its engine power. Well, that change is just given by the final velocity 𝑣𝑓 minus the initial velocity 𝑣𝑖. And that happens to be 11.5 minus 35, which in other words is negative 23.5 metres per second. We’re going to call this quantity Δ𝑣 because Δ means change. So we’re working out the change in velocity.

By the way, the value being negative does make sense because the train is losing speed, hence the negative value. So now that we’ve worked out the total change in speed, which was negative 23.5 metres per second, what we need to do is to work out how much speed it’s losing every second. We simply do this by dividing this change in speed by the number of seconds that there are. In other words, it’s just negative 23.5 divided by five.

This is because it’s losing 23.5 metres per second in five seconds. So in one second, it will lose 23.5 divided by five. And that turns out to be negative 4.7 metres per second squared.

By the way, this also deals with the problem of working out the average acceleration in the direction of motion, because in the direction of motion, the train is losing speed. So it’s decelerating or, in other words, its acceleration is negative. Hence, our final answer is that the train’s average acceleration in the direction of motion is negative 4.7 metres per second squared.

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