Answer:
Explanation:
Given the distance of the particle from the starting point at time t is given by the function: s=t⁵−6t⁴
velocity v(t) = ds/dt
[tex]v(t) = 5t^{5-1}-4(6)t^{4-1}\\v(t) = 5t^4-24t^3\\[/tex]
Next is to get the acceleration function:
[tex]a(t) = 4(5)t^{4-1}-3(24)t^{3-1}\\a(t) = 20t^3-72t^2[/tex]
Next is to get the value of t at which the acceleration is equal to zero
[tex]a(t) = 20t^3-72t^2\\0 = 20t^3-72t^2\\20t^3-72t^2 = 0\\t^2(20t-72) = 0\\t^2 =0 \ and \ 20t-72 = 0[/tex]
Since t ≠ 0, hence;
20t -72 = 0
20t = 72
t = 72/20
t = 3.6secs
Hence the value of t (other than 0 ) at which the acceleration is equal to zero is 3.6secs
