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If displaceme...

Question

If displacement s at time t is s = $t^{3} - 3 t^{2} - 15 t + 12$ , then acceleration at time t = 1 sec is

6 u n i t s / s e c^{2}

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- 6 u n i t s / s e c^{2}

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4 u n i t s / s e c^{2}

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Solution

The correct option is A 0
Velocity , $V = \frac{d s}{d t} = 3 t^{2} - 6 t - 15$
Acceleration, $a = \frac{\partial ϑ}{\partial t} = 6 t - 6$ At $t = 1$
$a = 0 m / s^{2} \Rightarrow (C)$

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Similar questions

t = 0

\to u = 4^i + 3^j m / s e c

\to a = 6^i + 4^j m / s e c^{2}

t = 2 s e c

θ = ω_{0} t + \frac{1}{2} α t^{2}

ω_{0}

α

ω_{0} = 1 rad/sec, α = 1.5 {rad/sec}^{2}

(

rad/s)

t = 2 sec

S = t^{3} + 3 t^{2} + 2 t - 1

t = 4 s e c

2 + 12 K m / s

k

a = 2 t - 6

(m / s^{2})

t

t = 0, u = 2 m / s, x = 2 m

θ = ω_{0} t + \frac{1}{2} α t^{2}

ω_{0}

α

ω_{0} = 1 rad/sec, α = 1.5 {rad/sec}^{2}

(

rad/s)

t = 2 sec

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