Question

please solve it answer is also given pleaseeeeeeeeeee help me tomorrow i m having my exam please help

Answer 1

Take the derivative to find the expression for the velocity of the particle. Take the derivative again to find the expression for the acceleration. Set the velocity expression equal to zero and solve the resulting quadratic for s. Plug this value into the expression for the acceleration. You should get ±6cm/sec2

Explanation:

s=t3–6t2+9t–5

v=dsdt=3t2–12t+9

a=dvdt=d2sdt2=6t–12

v=3t2–12t+9=0

3(t2–4t+3)=0

(−1)⋅(−3)=3 and (−1)+(−3)=−4

so x−3, and x−1 are factors

3(t–3)(t−1)=0

we have roots t1=1 and t2=3

Plug these into a

a1=6(1)–12=−6

a2=6(3)–12=6