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A race car entering the curved part of the track drops its speed from 48.3 m/s to 39.6 m/ s in 2.59 s. if the radius of the curved part of the track is 360 m, calculate the magnitude of the total acceleration, in m/s2, of the race car just after it has begun to reduce its speed.

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Final answer:

The total acceleration of the race car is calculated to be 5.51 m/s², which combines both the tangential deceleration of -3.35 m/s² and the centripetal acceleration of 4.38 m/s².

Step-by-step explanation:

To calculate the magnitude of the total acceleration of the race car as it begins to reduce its speed on a curved part of the track, we must consider both the tangential acceleration caused by the change in speed and the centripetal acceleration due to the change in direction.

First, we determine the tangential acceleration (a_t), which is the rate of change of speed in the direction of motion. It is calculated using the formula:

a_t = (final speed - initial speed) / time

In our case, this would be:

a_t = (39.6 m/s - 48.3 m/s) / 2.59 s = -3.35 m/s²

Note that the negative sign indicates a deceleration.

Next, we calculate the centripetal acceleration (a_c), which is required to keep the car moving in a circle and is directed towards the center of the circle. It is given by the formula:

a_c = v² / r

Where v is the speed of the car at the moment considered and r is the radius of the curvature.

a_c = (39.6 m/s)² / 360 m = 4.38 m/s²

Finally, the total acceleration (a_total) is found by combining the tangential and centripetal accelerations, since they are perpendicular to each other:

a_total = √(a_t² + a_c²)

a_total = √((-3.35 m/s²)² + (4.38 m/s²)²)

a_total = √(11.2225 + 19.1844) m/s²

a_total = √(30.4069) m/s²

a_total = 5.51 m/s²

Therefore, the magnitude of the total acceleration of the race car just after it has begun to reduce its speed is 5.51 m/s².

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