Question

a particle moves in a straight line its position is given by x^=t^2+1. its acceleration at time t

Answer 1

Answer:

$\boxed{\mathfrak{ a = 2 \ m/s^2}}$

Explanation:

Relation between position (x) of particle w.r.t time (t) is given as:

x = t² + 1

Acceleration (a) of particle is equal to double differentiation of position-time relation.

$\sf \implies \bold{a = \frac{ {d}^{2}x}{d {t}^{2} } } \\ \\ \sf \implies a = \frac{d( \frac{dx}{dt}) }{dt} \\ \\ \sf \implies a = \frac{d( \dfrac{d}{dt}( {t}^{2} + 1)) }{dt} \\ \\ \sf \implies a = \frac{d}{dt} (2t) \\ \\ \sf \implies a = 2 \: m {s}^{ - 2}$

Additional information:

$\sf \frac{d {x}^{n} }{dx} = n {x}^{n - 1}$