Question Video: Finding the Expression of the Position of a Particle in Terms of Time given Its Initial Position and Velocity Expression | Nagwa Question Video: Finding the Expression of the Position of a Particle in Terms of Time given Its Initial Position and Velocity Expression | Nagwa

Question Video: Finding the Expression of the Position of a Particle in Terms of Time given Its Initial Position and Velocity Expression Mathematics • Third Year of Secondary School

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A particle is moving in a straight line such that its velocity at time π‘ seconds is given by π£ = [βsin (4π‘) + 14] m/s, π‘ β₯ 0. Given that its initial position πβ = 13 m, find an expression for its position at time π‘ seconds.

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Video Transcript

A particle is moving in a straight line such that its velocity after π‘ seconds is given by π£ is equal to negative sin of four π‘ plus 14 meters per second, where π‘ is greater than or equal to zero. Given that its initial position π zero is equal to 13 meters, find an expression for its position at time π‘ seconds.

The question tells us a particle is moving in a straight line such that its velocity after π‘ seconds is given by negative the sin of four π‘ plus 14 meters per second. And weβre told this is only valid for values of π‘ greater than or equal to zero. Weβre also given the initial position of our particle π zero is 13 meters. The question wants us to use this information to find an expression for the position of our particle after π‘ seconds.

Since weβre given the velocity of our particle, we can recall that the velocity of our particle is actually the rate of change in displacement with respect to time. Velocity is the derivative of π  with respect to π‘. However, the converse to this statement is also true. The displacement of our particle after π‘ seconds is an antiderivative of the velocity with respect to time. And since our particle is moving in a straight line, we can measure its position by measuring the displacement. So if we integrate negative sin of four π‘ plus 14 with respect to π‘, weβll get an expression for the displacement function up to a constant of integration.

To evaluate this integral, we recall for constants π and π, where π is not equal to zero, the integral of π times the sin of ππ‘ with respect to π‘ is equal to negative π times the cos of ππ‘ over π plus a constant of integration πΆ. We can then use this to evaluate our integral. We get negative one multiplied by negative the cos of four π‘ divided by four. We recall the antiderivative of a constant is just equal to that constant multiplied by π‘. So we integrate 14 to get 14π‘. And then we add a constant of integration we will call πΆ.

We can simplify this expression. We know that negative one multiplied by negative one is just equal to one. So weβve shown the displacement of our particle after time π‘ is given by the cos of four π‘ divided by four plus 14π‘ plus πΆ. We call this a general solution because we donβt know the value of πΆ. To help us find the value of πΆ, we recall that the initial position of our particle, π zero, is equal to 13. Substituting π‘ is equal to zero into our equation for the displacement gives us π  evaluated at zero is equal to the cos of four times zero over four plus 14 times zero plus πΆ. And π  evaluated at zero would be the displacement of our particle after zero seconds. It would be the initial displacement of our particle. And we know that this is equal to 13.

We can simplify this expression. 14 times zero is equal to zero, and four times zero is equal to zero. And we know the cos of zero is equal to one. So we now have our initial displacement of 13 is equal to one-quarter plus πΆ. Subtracting one-quarter from both sides of this equation, we see that πΆ is equal to 13 minus one-quarter. And we can calculate this to give us 51 divided by four. So weβve shown that πΆ is equal to 51 divided by four. We can substitute this into our equation for the displacement of our particle after π‘ seconds. And this gives us that π  of π‘ is equal to 14π‘ plus one-quarter times the cos of four π‘ plus 51 divided by four.

And since all of our units are given in terms of meters and seconds and this represents the displacement after π‘ seconds, we can add the units of meters. Therefore, weβve shown if a particle is moving in a straight line where its velocity after π‘ seconds is given by negative sin of four π‘ plus 14 meters per second where π‘ is greater than or equal to zero and its initial position is 13 meters. Then the displacement of the particle after π‘ seconds is given by 14π‘ plus one-quarter times the cos of four π‘ plus 51 over four meters.

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