Suppose that a particle moves according to the law of motion...

Suppose that a particle moves according to the law of motion s(t) = t^4 - 6t^3 + 4t^2 + 3 with t ? 0, where s is in meters and t is in seconds. Use your knowledge of the relationship between position, velocity, and acceleration to answer the following questions. Use exact values (not approximations) throughout. a) Find the velocity of the particle at time t. What is the velocity at t = 0? b) Determine, using algebraic analysis, when the particle is moving in a positive direction, a negative direction, and when it is at rest. c) Find the total distance traveled by the particle in the first 6 seconds. d) Find the acceleration of the particle at time t. What is the acceleration at t = 0? e) When is the particle speeding up and slowing down? Explain how you found your answers.

Transcript

00:01 Hello students, in this question s of t is given, s of t is equal to t raised to 4 minus 6t raised to cube plus 4t raised to square plus 3.

00:16 So, in the a part of the question we have to find the velocity v at t seconds and at t is equal to 0.

00:25 So, for that first we have to take the derivative of s of t that is velocity v is equal to d by dt of s of t.

00:34 So, taking the derivative we get d by dt of substituting s of t that is t raised to 4 minus 6t raised to cube plus 4t square plus 3.

00:50 Now, let us take the derivative of this that is 4t raised to 3 minus 6 into 3t square plus 4 into 2t plus 0 and this is equal to 4t cube minus 18t square plus 8t.

01:18 So, this is the velocity v.

01:22 Now, we have to find velocity v this is v of t.

01:26 Now, we have to find velocity at time t is equal to 0 that is v of 0 and this is given by 0 minus 0 plus 0 that is v of 0 is equal to 0.

01:43 So, this is the answer to the a part of the question.

01:48 Now, in the b part of the question we are asked to find out when will the particle be moving in a positive direction a negative direction and when will it come to rest.

01:57 For that we can consider the equation for velocity v is equal to 4t cube minus 18t square plus 8t.

02:08 Now, if we take t outside we can write this as 4t square minus minus 18t plus 8.

02:22 Now, we know when t is equal to 0 v of t is also equal to 0 and equating this to 0 we get 4t square minus 18t plus 8 is equal to 0.

02:38 On simplifying we get we can write it as 2t square minus 8t minus 1t plus 4 is equal to 0.

02:52 2t t minus 4 now taking this together we can write it as minus outside t minus 4 is equal to 0 that is t minus 4 2t minus 1 is equal to 0 that is t is equal to 4 and we get t is equal to 1 by 2...

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