Question Video: Acceleration over Distance and Time | Nagwa Question Video: Acceleration over Distance and Time | Nagwa

Question Video: Acceleration over Distance and Time Physics • First Year of Secondary School

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A bicycle has an initial velocity of 12 m/s and it accelerates for 15 s in the direction of its velocity, moving 220 m during that time. What is the bicycle’s rate of acceleration? Answer to two decimal places.

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Video Transcript

A bicycle has an initial velocity of 12 meters per second. And it accelerates for 15 seconds in the direction of its velocity, moving 220 meters during that time. What is the bicycle’s rate of acceleration? Answer to two decimal places.

Alright, so in this question, we’ve got a bicycle. So let’s start by saying that it’s moving in this direction. We choose this arbitrarily. And we’ve been told that this bicycle has an initial velocity of 12 meters per second. Now, we’ve been told that this bicycle accelerates for 15 seconds in the direction of its velocity. In other words, it accelerates for 15 seconds in this direction, the same direction that it was initially travelling in. So its velocity is increasing. It’s getting faster and faster. So here’s our bicycle after accelerating. And we’ve been told that the acceleration phase takes a time of 15 seconds. And in that time, we’ve been told that the bicycle moves a distance of 220 meters.

Now, based on all of this information, we need to find the rate of the bicycle’s acceleration. So if we say that what we’re trying to find, the acceleration, is 𝑎. And if we further say that the initial velocity of the bicycle was 𝑢, the time taken to accelerate was 𝑡. And the distance moved by the bicycle was 𝑠. Then we need to find an equation that links together the quantities 𝑢, 𝑡, 𝑠, and 𝑎.

Now, the equation that we’re looking for is one of the kinematic equations. Specifically, the equation that tells us that the distance moved by the bicycle in a straight line or, in other words, the displacement of the bicycle. Is equal to the initial velocity of the bicycle multiplied by the time over which the bicycle accelerates plus half multiplied by the acceleration multiplied by the time taken for the acceleration squared. In other words, 𝑠 is equal to 𝑢𝑡 plus half 𝑎𝑡 squared.

Now, in this equation, we know the value of 𝑠, the value of 𝑢, and the value of 𝑡. And the only thing we don’t know is the value of 𝑎. So we need to rearrange and solve for 𝑎. To do this, we can start by subtracting 𝑢𝑡 from both sides of the equation. This way, on the right-hand side, 𝑢𝑡 cancels with minus 𝑢𝑡. And so what we’re left with is 𝑠 minus 𝑢𝑡 is equal to half 𝑎𝑡 squared. Then, we can multiply both sides of the equation by two divided by 𝑡 squared. And this way, the factor of half on the right-hand side cancels with the factor of two in the numerator. And we have 𝑡 squared divided by 𝑡 squared. Those cancel too, leaving us with just 𝑎 on the right-hand side.

And so when we finally rearrange the equation, we find that the acceleration of the bicycle is equal to two multiplied by 𝑠 minus 𝑢𝑡, all divided by 𝑡 squared. Then, we just need to plug in some values. We could say that the acceleration of the bicycle 𝑎 is equal to two multiplied by 𝑠, which is 220 meters, minus 𝑢𝑡. So we plug in 𝑢, which is 12 meters per second, and 𝑡, which is 15 seconds. And then we divide this whole thing by 𝑡 squared, which happens to be 15 seconds whole squared. Now when it comes to units, everything might look a little bit messy. We’ve got metres here. We’ve got meters per second here, seconds here, and second squared in the denominator. However, things are gonna cancel out very nicely. And to see that, let’s start by simplifying this pair of parentheses here.

What we’ve got is 12 meters per second multiplied by 15 seconds. So to calculate that, we first need to calculate 12 multiplied by 15 to give us the numerical value. And then we need to calculate meters divided by seconds multiplied by seconds. And so when we put the whole thing together, the numerical value, 12 times 15, ends up being 180. And then we see that this factor of seconds in the numerator cancels with this factor of seconds in the denominator. And what we’re left with is simply meters.

Hence, we can replace the power of parentheses with 180 meters. Then we can see that all we have in the numerator of this large fraction now is the unit of meters in both case. We’re subtracting 180 meters from 220 meters. And then we multiply all of that simply by a number, which is two in this case. So simplifying the whole of the numerator, we have 220 meters minus 180 meters, which is 40 meters. And then we multiply 40 meters by two, which is 80 meters. And now our fraction is looking a lot nicer.

To make it look even more clean, let’s simplify this parenthesis. We’ve got 15 seconds whole squared. So to work that out, we need to square the 15. And we need to square the seconds. So 15 squared is 225. And second squared is well second squared. Then we can see that our fraction is going to have a numerical value of 80 divided by 225. And a unit of meters divided by second squared or meters per second squared. Which is perfect because we are trying to calculate an acceleration here. And the base unit of acceleration is meters per second squared.

So all that’s left to do now is to calculate the numerical value 80 divided by 225 to two decimal places. When we do this, we find that the acceleration of the bicycle is 0.36 meters per second squared. And so at this point, we found the answer to our question.

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