Which of the following relations representing displacement x of a particle describes motion with constant acceleration?

x = 6 - 7 t^{2}

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x = 3 t^{2} + 5 t^{3} + 7

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x = 9 t^{- 2} + 8

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x = 4 t^{- 2} + 3 t^{- 1}

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Solution

The correct option is A $x = 6 - 7 t^{2}$
Constant acceleration means $a = \frac{d^{2} x}{d t^{2}}$ should be $i n d e p e n d e n t$ from $t$
so only $x = 6 - 7 t^{2}$ will follow this and other will not because other have the power of $t$ other than $2$ so on $d o u b l e$ differentiating
w.r.t. $t$ will not result in vanishing of $t$ .

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Similar questions

When we divide $3 t^{4} + 5 t^{3} - 7 t^{2} + 2 t + 2$ by $t^{2} + 3 t + 1$ , we get $3 t^{2} - 4 t + 2$ .

When we divide $3 t^{4} + 5 t^{3} - 7 t^{2} + 2 t + 2$ by $t^{2} + 3 t + 1$ , we get $3 t^{2} - 4 t + 2$ as quotient.

x = t + 2 t^{2} + 3 t^{3} - 4 t^{4}

a = 4 t + 3

(m / s^{2})

t

t = 0, u = 3 m / s, x = 0 m

t

x = 2 + 5 t + 7 t^{2}

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