Question Video: Finding the Acceleration of an Object over a Distance and Time | Nagwa Question Video: Finding the Acceleration of an Object over a Distance and Time | Nagwa

Question Video: Finding the Acceleration of an Object over a Distance and Time Physics • First Year of Secondary School

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A bicycle has an initial velocity of 10 m/s and it accelerates for 7.5 seconds. After this time, the bicycle is at a distance of 55 m from the point at which it began accelerating. What was the bicycle’s rate of acceleration? Answer to two decimal places.

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Video Transcript

A bicycle has an initial velocity of 10 meters per second and it accelerates for 7.5 seconds. After this time, the bicycle is at a distance of 55 meters from the point at which it began accelerating. What was the bicycle’s rate of acceleration? Answer to two decimal places.

In this question, we’re being asked about the motion of a bicycle. We’re told that the bicycle has an initial velocity of 10 meters per second, and we’ll label this initial velocity as 𝑢. Let’s recall that velocity is a vector quantity, which means that it has both a magnitude and a direction. The question gave us the magnitude of this initial velocity as 10 meters per second, but it didn’t say anything about the direction.

In our sketch though, we’ve chosen to draw this bicycle moving towards the right, so let’s add an arrow pointing to the right to represent the direction of this initial velocity. Since we’ve got this initial velocity directed to the right and it has a positive value, then that means that right is our positive direction. In other words, quantities directed to the right will have positive values while those directed to the left will have negative values.

We’re also told in the question that starting from this initial velocity, the bicycle accelerates for a time of 7.5 seconds. Let’s label this time as 𝑡. After this time 𝑡 has passed, we know that the bike is at a distance of 55 meters from the point at which it started accelerating. We could also say that the bike is displaced in the positive direction by 55 meters after this time 𝑡. We’ve labeled this displacement as 𝑠.

Given all of this information, we’re asked to work out what the bicycle’s acceleration is. Let’s label this acceleration as 𝑎. And we know that if the value we find for 𝑎 is positive, then the acceleration is directed to the right. Meanwhile, if we find a negative value for 𝑎, then that means that the acceleration is directed to the left. While a positive value of acceleration would mean that the bike is speeding up, increasing its velocity, a negative acceleration, or deceleration, would mean that the bike is slowing down, decreasing its velocity in the positive direction.

In order to work out the value of the acceleration 𝑎, we’re going to need to recall one of the kinematic equations of motion. Specifically, the equation that we want is this one here, which says that 𝑠 is equal to 𝑢 times 𝑡 plus a half times 𝑎 times 𝑡 squared. In this equation, all of the letters have the same meaning as those we’ve assigned to the quantities given to us in the question. That is, 𝑠 represents the displacement of an object, 𝑢 is the object’s initial velocity, 𝑎 is the acceleration of the object, and 𝑡 is the time for which it accelerates.

Applying this kinematic equation to the bicycle from this question, we know the value of 𝑠, we know the value of 𝑢, and we know the value of 𝑡. The quantity that we don’t know and that we want to work out is the bicycle’s acceleration 𝑎. If we rearrange the equation in order to make 𝑎 the subject, then we’ll be able to use it to solve our problem. First, though, we should note that this equation comes with a couple of caveats. In particular, the equation only applies when the acceleration 𝑎 is constant and the motion is in a straight line.

Since we’re asked to calculate a single value for the acceleration of the bike, then we can assume that this acceleration 𝑎 is indeed constant because otherwise there wouldn’t be just a single value; the acceleration would be changing in time. Additionally, by equating the distance of 55 meters to a displacement of 55 meters, we have already implicitly assumed that the motion is in a straight line. Since there’s no mention in the question of any turning or changing direction of the bicycle, then it’s fair to assume that it does indeed travel in a straight line. Since these two conditions are then met, that means that we can safely use this equation in order to work out the acceleration 𝑎.

So then let’s get on with rearranging this equation in order to make 𝑎 the subject. We’ll begin by subtracting 𝑢 times 𝑡 from both sides. Because this means that on the right, the positive and negative 𝑢 times 𝑡 terms cancel out. We have then that 𝑠 minus 𝑢 times 𝑡 is equal to a half times 𝑎 times 𝑡 squared. Then, we’ll multiply both sides of this by two over 𝑡 squared. This means that on the right-hand side the factors of two in the numerator and denominator cancel each other out. And likewise, the factors of 𝑡 squared also cancel. The only quantity then remaining on the right-hand side is the acceleration 𝑎. If we write the equation the other way around, we have that 𝑎 is equal to two multiplied by 𝑠 minus 𝑢 times 𝑡 all divided by 𝑡 squared.

Let’s now clear ourselves some space on the board so that we can substitute our values into the right-hand side of this equation. Substituting in that the displacement 𝑠 is equal to 55 meters, the initial velocity 𝑢 is 10 meters per second, and the time 𝑡 is 7.5 seconds, we end up with this expression for the acceleration 𝑎. Let’s begin by considering the numerator of this expression.

Inside this outer set of parentheses, we’ve got 55 meters minus 10 meters per second multiplied by 7.5 seconds. Notice that in this right-hand term, the units of seconds cancel with the units of per second. This means that the right-hand term overall then has units of meters, which matches the units of meters from the left-hand term. So, we can rewrite this right-hand term as 10 multiplied by 7.5 with units of meters. Evaluating this product of 10 times 7.5 gives a result of 75. Then, we’ve got 55 meters minus 75 meters, and this works out as negative 20 meters. That means that the numerator of this expression is equal to two multiplied by negative 20 meters, which evaluates to negative 40 meters.

If we now look at the denominator of the expression, we can see that this is equal to the square of 7.5 seconds. We can equivalently write that as 7.5 squared with units of second squared. Then, the square of 7.5 works out as 56.25. So, we’ve got that the acceleration 𝑎 is equal to negative 40 meters divided by 56.25 seconds squared. Grouping the units together on the right-hand side gives us 𝑎 is equal to negative 40 over 56.25 with units of meters per second squared. When we do this division, we get a result for the acceleration 𝑎 of negative 0.71 recurring meters per second squared.

Notice that since this value is negative, then the acceleration is in the opposite direction to the bicycle’s initial velocity. So, this acceleration is acting to slow the bike down. Now, we just have one remaining step because the question wants our answer given to two decimal places. When we round negative 0.71 recurring to two decimal places, we get negative 0.71.

Then, our answer to this question is that, to two decimal places, the acceleration of the bicycle is equal to negative 0.71 meters per second squared.

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