Question

An object moves in a circle of radius 29 m with its speed given by v = 3.6 + 1.5t2 , with v in meters per second and t in seconds. At t = 3.5 s Find the tangential acceleration.

Answer 1

The tangential acceleration of the object at t = 3.5 s is 10.5 m/s^2.

Step-by-step explanation:

To find the tangential acceleration, we first need to find the velocity of the object at time t = 3.5 seconds. We can do this by substituting t = 3.5 into the given expression for v:

v = 3.6 + 1.5t^2

v = 3.6 + 1.5(3.5)^2

v = 25.725 m/s

The tangential acceleration at time t = 3.5 seconds is given by the derivative of the velocity with respect to time:

a_t = dv/dt = 3.0t

Substituting t = 3.5 seconds into this expression, we get:

a_t = 3.0(3.5) = 10.5 m/s^2

Therefore, the tangential acceleration of the object at t = 3.5 s is 10.5 m/s^2.