A point is moving along a curve so that it is a distances =3t² + 2t m from its starting point, where t is in seconds. At t =2 s the point is at a location where the curve has a radius of curvature rho = 3m.
What is the magnitude of the acceleration at this time?

The magnitude of the acceleration at 2 seconds is 65.33 m/s²

Explanation:

Given;

distance of the point along a curve = (3t² + 2t) m

let this distance = y

Then, velocity = dy/dt

y = 3t² + 2t

dy/dt = 6t + 2

magnitude of velocity at t = 2 s, = 6(2) + 2 = 12 + 2 = 14 m/s

centripetal acceleration, a = v²/r

where;

a is the magnitude of the centripetal acceleration

v is the magnitude of the velocity

r is the radius of the curve

a = v²/r

a = (14)²/3

a = 65.33 m/s²

Therefore, the magnitude of the acceleration at 2 seconds is 65.33 m/s²

step-by-step explanation:

we are given

we have to simplify it

firstly, we will factor it

now, we can cancel like terms

now, we can combine it

so, we get

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