Final answer:
The tangential component of acceleration at t = 3 is 3000 ft/s² and the normal component of acceleration is 81000 ft/s².
Step-by-step explanation:
To find the tangential and normal components of acceleration, we need to determine the total acceleration and then resolve it into its tangential and radial components.
The tangential component of acceleration is the rate of change of speed along the circular path, while the normal component of acceleration is the centripetal acceleration responsible for keeping the car on the circular path.
Given that the radius of the circular path is 1000 ft, the tangential acceleration can be found using the formula:
at = r * α
where at is the tangential acceleration, r is the radius, and α is the angular acceleration. We are given that the car's speed increases at a rate of t ft/s², so at t = 3, the tangential acceleration is at = r * t. Substituting the values, we have:
at = 1000 ft * 3 ft/s² = 3000 ft/s²
The normal component of acceleration is the centripetal acceleration, given by the formula:
ac = v² / r
where ac is the centripetal acceleration and v is the velocity. Since the car starts at rest, the velocity at t = 3 is v = r * α * t. Substituting the values, we have:
v = 1000 ft * 3 ft/s² * 3 s = 9000 ft/s
Plugging this value into the formula for centripetal acceleration, we get:
ac = (9000 ft/s)² / 1000 ft = 81000 ft/s²
Therefore, at t = 3, the tangential component of acceleration is 3000 ft/s² and the normal component of acceleration is 81000 ft/s².