Question Video: Calculating the Acceleration of an Object over a Distance | Nagwa Question Video: Calculating the Acceleration of an Object over a Distance | Nagwa

# Question Video: Calculating the Acceleration of an Object over a Distance Physics • First Year of Secondary School

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An object starts from rest and accelerates along an 8 m long straight line. Its velocity reaches 12 m/s when it is at the end of the line. What is the object’s acceleration along the line?

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### Video Transcript

An object starts from rest and accelerates along an eight-meter-long straight line. Its velocity reaches 12 meters per second when it is at the end of the line. What is the object’s acceleration along the line?

So, this question asks us to find the acceleration of an object. We’re told that the object starts from rest. In this question, the phrase “from rest” means that the object is initially not moving. Therefore, its initial velocity is zero meters per second. The question then tells us that the object reaches a velocity of 12 meters per second after moving from rest along an eight-meter-long straight line.

Recall that acceleration is defined as the rate of change of velocity. This increase in velocity from zero meters per second to 12 meters per second tells us that the object has accelerated along the line. Now recall that velocity and acceleration are vector quantities, meaning that they have a direction as well as a magnitude.

The question tells us that the object is moving along a straight line. This means that it doesn’t change direction, so we know for sure that the acceleration’s direction is constant. Since we’re asked to calculate a single value for this acceleration, we can assume that the magnitude of the acceleration is constant too.

To help us answer the question, let’s use symbols to rewrite the information that we’ve been given. We want to find the object’s acceleration, and we can label this as 𝑎. At the moment, the value of 𝑎 is unknown. Next, we can label the initial velocity as 𝑢, which we’ve learnt is equal to zero meters per second. The final velocity can be labeled as 𝑣, and this is equal to the 12 meters per second that the object reaches at the end of the line. Lastly, we can label the distance traveled along the line as 𝑠, and this is equal to eight meters.

Using this information, as well as remembering that the object’s acceleration will be constant, we can use an equation of motion, also known as a kinematic equation, to help us calculate the answer. The kinematic equation that we’ll want to use is 𝑣 squared is equal to 𝑢 squared plus two times 𝑎 times 𝑠. We are solving for the acceleration 𝑎. This means we need to rearrange the equation in order to make 𝑎 the subject.

First, we can subtract 𝑢 squared from both sides. After canceling out the positive and negative 𝑢 squared terms on the right-hand side, we are now left with 𝑣 squared minus 𝑢 squared equals two times 𝑎 times 𝑠. Let’s now divide both sides by two 𝑠. After canceling out two 𝑠 from the numerator and denominator on the right-hand side, we are finally left with the equation that we can use to find the acceleration. 𝑣 squared minus 𝑢 squared divided by two 𝑠 is equal to the acceleration 𝑎.

We can now substitute into the equation the values that we have for 𝑣, 𝑢, and 𝑠. This gives us that the acceleration 𝑎 is equal to the square of 12 meters per second minus the square of zero meters per second all divided by two times eight meters.

Notice that the velocities in the numerator each have units of meters per second. Taking the square of these units gives units of meters squared per second squared. We can then cancel one factor of meters from the terms in the numerator with the units of meters from the denominator. This gives us overall units for the acceleration of meters per second squared. When we type this expression into a calculator, we get a value for this acceleration of nine meters per second squared.

Therefore, our final answer is that the object’s acceleration along the line is nine meters per second squared.

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