Final answer:
The total acceleration of a 1000-kg car moving at 16.8 m/s around a circular turn of 18.5 m radius with a deceleration of 35.2 m/s² is approximately 38.4 m/s².
Step-by-step explanation:
Total Acceleration of a Car in Circular Motion
A 1000-kg car is decreasing in speed at a rate of 35.2 m/s² while moving around a circular turn with a radius of 18.5 meters. At the instant the car is moving at 16.8 m/s, we need to calculate the car's total acceleration. This involves two components: the tangential deceleration ('rate of change of the speed') and the centripetal acceleration ('acceleration towards the center of the circular path').
The tangential acceleration is already given as 35.2 m/s². However, this is deceleration, its direction is opposite to the motion of the car, so we take it as -35.2 m/s².
To find the centripetal acceleration (ac), we use the formula:
ac = v² / r
where v is the speed (16.8 m/s), and r is the radius (18.5 m). Plugging the numbers we get:
ac = (16.8 m/s)² / 18.5 m ≈ 15.3 m/s²
Now we have to combine these two accelerations to get the total acceleration. Since they are perpendicular to each other (tangential is along the motion, centripetal is towards the center of the circle), we add them as vectors:
Total acceleration = √((-35.2)^2 + (15.3)^2) m/s² ≈ √(1241.44 + 234.09) m/s² ≈ √(1475.53) m/s² ≈ 38.4 m/s²
This is the magnitude of the total acceleration of the car at the instant.