Question

S=t³+2t²+3t+4,find acceleration
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Answer 1

Answer:

6t+4

Explanation:

$s = {t}^{3} + 2 {t}^{2} + 3t + 4$

velocity=ds/dt

$v = \frac{ds}{dt} = \frac{d( {t}^{3} + 2 {t}^{2} + 3t + 4) }{dt}$

$v = 3 {t}^{2} + 4t + 3$

acceleration=dv/dt

$a = \frac{dv}{dt} = \frac{d(3 {t}^{2} + 4t + 3) }{dt}$

$a = 6t + 4$

so answer is 6t+4