Video Transcript
A particle was moving in a straight line with a constant acceleration. If the particle covered 17 meters in the second second and 46 meters in the ninth and 10th seconds, calculate its acceleration 𝑎 and its initial velocity 𝑣 sub zero.
In order to answer this question, we will use the equations of motion or SUVAT equations, where 𝑠 is the displacement measured in standard units of meters, 𝑢 and 𝑣 are the initial and final velocities measured in meters per second, 𝑎 is the acceleration in meters per second squared, and 𝑡 is the time measured in seconds. In the question, we are given two pieces of information, firstly that the particle covers 17 meters in the second second. We are also told that it covers 46 meters in the ninth and 10th seconds. As the constant acceleration of the particle is 𝑎, the velocity of the particle will increase by 𝑎 meters per second for each second of motion.
Let’s consider our first piece of information. We know that the time here is one second and the displacement is 17 meters. We have the constant acceleration of 𝑎 meters per second squared. Since the initial velocity of the particle is 𝑣 sub zero, after one second, it will be traveling with velocity 𝑣 sub zero plus 𝑎. Therefore, our value of 𝑢 here is 𝑣 sub zero plus 𝑎 meters per second. The final velocity one second later will be 𝑣 sub zero plus two 𝑎 meters per second. Next, we can substitute our values of 𝑠, 𝑢, 𝑎, and 𝑡 into the equation. 𝑠 is equal to 𝑢𝑡 plus a half 𝑎𝑡 squared. This gives us 17 is equal to 𝑣 sub zero plus 𝑎 multiplied by one plus a half multiplied by 𝑎 multiplied by one squared.
The right-hand side simplifies to 𝑣 sub zero plus 𝑎 plus a half 𝑎, which in turn becomes 𝑣 sub zero plus three over two 𝑎. Multiplying through by two, we get 34 is equal to two 𝑣 sub zero plus three 𝑎. As there are two unknowns here, we will call this equation one. In the second piece of information about the particle, we have a displacement of 46 meters and a time of two seconds. Once again, the acceleration is 𝑎 meters per second squared. As we are dealing with the ninth and 10th seconds and we recall the velocity will increase by 𝑎 meters per second for each second of motion, 𝑢 is equal to 𝑣 sub zero plus eight 𝑎 meters per second and 𝑣 is equal to 𝑣 sub zero plus 10𝑎 meters per second.
After clearing some space, we can use the same equation to consider this period of the particle’s motion. This time, we have 46 is equal to 𝑣 sub zero plus eight 𝑎 multiplied by two plus a half multiplied by 𝑎 multiplied by two squared. This time, the right-hand side simplifies to two 𝑣 sub zero plus 16𝑎 plus two 𝑎. And collecting like terms, we have 46 is equal to two 𝑣 sub zero plus 18𝑎. We will call this equation two. We now have a pair of simultaneous equations that we can solve by elimination to calculate the values of 𝑎 and 𝑣 sub zero.
We begin by subtracting equation one from equation two. 46 minus 34 is equal to 12. Two 𝑣 sub zero minus two 𝑣 sub zero is equal to zero, and 18𝑎 minus three 𝑎 is 15𝑎. Dividing both sides of this equation by 15, we have 𝑎 is equal to 12 over 15. This fraction can be simplified to four-fifths or 0.8. The acceleration 𝑎 is therefore equal to 0.8 meters per second squared. We can now substitute this value of 𝑎 back into equation one. This gives us 34 is equal to two 𝑣 sub zero plus three multiplied by 0.8. Three multiplied by 0.8 is 2.4. And we can subtract this from both sides of the equation. This gives us two 𝑣 sub zero is equal to 31.6. And dividing by two, 𝑣 sub zero is equal to 15.8. The initial velocity of the particle 𝑣 sub zero is equal to 15.8 meters per second. We now have the two answers required in this question.
