aphathalo

2022-10-06

The displacement of a car is given by the formula $s=-2{t}^{3}+2{t}^{2}+16t-1$ where s is in meters and t is in second.

4.1 Find the formula for the velociry of the car at any time t.

4.2 Find the acceleration of the car when t=5 seconds.

4.1 Find the formula for the velociry of the car at any time t.

4.2 Find the acceleration of the car when t=5 seconds.

Allvin03

Beginner2022-10-07Added 7 answers

Given

$s=-2{t}^{3}+2{t}^{2}+16t-1$

a) $v(t)={s}^{\prime}(t)=-3(2{t}^{2})+2(2t)+16(1)\phantom{\rule{0ex}{0ex}}v(t)=-6{t}^{2}+4t+16$

b) ^2a(t)=s''(t)=-2(6t)+4(1)\=[-12t+4]\a(5)=-12\times5+4=-60+4=-54m/s^2Show External URL Show Embeded Code Hide MathML Code a(t)=s′′(t)=−2(6t)+4(1)=[−12t+4]a(5)=−12×5+4=−60+4=−54m/s2$a(t)={s}^{\u2033}(t)=-2(6t)+4(1)\phantom{\rule{0ex}{0ex}}=[-12t+4]\phantom{\rule{0ex}{0ex}}a(5)=-12\times 5+4=-60+4=-54m/{s}^{2}$

$s=-2{t}^{3}+2{t}^{2}+16t-1$

a) $v(t)={s}^{\prime}(t)=-3(2{t}^{2})+2(2t)+16(1)\phantom{\rule{0ex}{0ex}}v(t)=-6{t}^{2}+4t+16$

b) ^2a(t)=s''(t)=-2(6t)+4(1)\=[-12t+4]\a(5)=-12\times5+4=-60+4=-54m/s^2Show External URL Show Embeded Code Hide MathML Code a(t)=s′′(t)=−2(6t)+4(1)=[−12t+4]a(5)=−12×5+4=−60+4=−54m/s2$a(t)={s}^{\u2033}(t)=-2(6t)+4(1)\phantom{\rule{0ex}{0ex}}=[-12t+4]\phantom{\rule{0ex}{0ex}}a(5)=-12\times 5+4=-60+4=-54m/{s}^{2}$

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