Video: Finding the Uniform Acceleration of a Body and Its Initial Velocity

A body was uniformly accelerating in a straight line such that it covered 72 m in the first 3 seconds and 52 m in the next 4 seconds. Find its acceleration π‘Ž and its initial velocity 𝑣₀.

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Video Transcript

A body was uniformly accelerating in a straight line such that it covered 72 metres in the first three seconds and 52 metres in the next four seconds. Find its acceleration π‘Ž and its initial velocity 𝑣 nought.

In order to solve this problem, we’ll use one of our equations of motion or π‘ π‘’π‘£π‘Žπ‘‘ equations. 𝑠 is equal to 𝑒 multiplied by 𝑑 plus a half π‘Žπ‘‘ squared, where 𝑠 is the displacement, 𝑒 is the initial velocity, 𝑣 is the final velocity, π‘Ž is the acceleration, and 𝑑 is equal to the time.

Our journey in the question has been split into two parts: from point 𝐴 to point 𝐡, we have travelled 72 metres in three seconds and from point 𝐡 to point 𝐢, we have travelled 52 metres in four seconds. The acceleration throughout this journey was constant or uniform.

If we consider the first part of the journey from 𝐴 to 𝐡, our 𝑠 or displacement is 72 metres, our time is three seconds, the acceleration is the unknown π‘Ž, and the initial velocity is the unknown 𝑣 zero. Substituting these values into the equation 𝑠 equals 𝑒𝑑 plus half π‘Žπ‘‘ squared gives us 72 is equal to three multiplied by 𝑣 zero plus 4.5π‘Ž. Dividing both sides of this equation by three gives us 24 is equal to 𝑣 zero plus 1.5π‘Ž. We will call this equation number one.

If we now consider the entire journey from point 𝐴 to point 𝐢, we can see that the total distance or displacement is 124 metres as 72 plus 52 is 124. The total time is seven seconds, three plus four is equal to seven. As the acceleration was uniform, this is still π‘Ž and the initial velocity is 𝑣 nought or 𝑣 zero.

Substituting these values into the equation 𝑠 equals 𝑒𝑑 plus half π‘Žπ‘‘ squared gives us 124 is equal to seven 𝑣 zero plus 24.5π‘Ž. Dividing both sides of this equation by seven gives us a hundred and twenty-four sevenths is equal to 𝑣 zero plus 3.5π‘Ž. We shall call this equation number two.

We now have two simultaneous equations, which we can solve to work out the acceleration π‘Ž and the initial velocity 𝑣 zero. The two equations are 24 equals 𝑣 zero plus 1.5π‘Ž and one hundred and twenty-four sevenths equals zero plus 3.5π‘Ž. Subtracting equation one from equation two gives us negative forty-four sevenths is equal to 2π‘Ž. Dividing both sides of this equation by two gives us a value for π‘Ž of negative twenty-two sevenths. Therefore, the uniform acceleration is negative twenty-two sevenths metres per second squared.

In order to calculate 𝑣 zero, the initial velocity, we need to substitute this value of π‘Ž, negative twenty-two sevenths, into one of the equations. In this case, we’ll substitute it into number two. This gives us one hundred and twenty-four sevenths equals 𝑣 zero minus 11. Adding 11 to both sides of this equation gives us a value of 𝑣 zero of two hundred and one sevenths.

Therefore, the initial velocity is two hundred and one sevenths metres per second or 28.71 metres per second. We could then use this information, the initial velocity and the acceleration, to work out the velocity at point 𝐡 and also at point 𝐢.

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