Answer:
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
