If a particle moves along a straight line according to s(t)= t^4 -4t^3 +6t^2 -20, find the maximum and minimum acceleration on 0 less or equal to t less or equal to 3

so we have s(t), namely the positional equation for the particle.

now, let's recall that the derivative of s(t) will be v(t), namely the derivative of the positional equation will be the velocity, and the derivative of the velocity will be the acceleration.

s'(t) = v(t) and s''(t) = a'(t)  <--- acceleration

so we need the 2nd derivative of s(t) firstly, and then we can check for any critical values and find its extrema.

now, if we run a first-derivative test on that acceleration derivative, namely s'''(t), let's say we test s'''(0) and that gives us a negative value, and then test s'''(2) and that gives us a positive value, so the test goes like in the picture below, clearly, 1 = t is a minumum, and is between [0, 3].

s'''(0) = -24

s'''(3) =  48

so clearly the maximum happens at the s'''(3) endpoint, when t = 3.

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐

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