Acar is moving with a speed of 32.0 m/s. the driver sees an accident ahead and slams on the brakes, causing the car to slow down with a uniform acceleration of magnitude 3.50 m/s2. how far does the car travel after the driver put on the brakes until it comes to a stop? a car is moving with a speed of 32.0 m/s. the driver sees an accident ahead and slams on the brakes, causing the car to slow down with a uniform acceleration of magnitude 3.50 m/s2. how far does the car travel after the driver put on the brakes until it comes to a stop? 112 m 292 m 4.57 m 146 m 9.14 m

The car will travel up to a distance of 146.28 m, hence the correct answer is 146 m.

Since the car is decelerating with the constant acceleration, so we can apply the third equation of motion.

v^2=u^2-2aS

here, v is the final speed of the car, which is 0 as the car stops,

u is the initial speed=32 m/s

a is the constant acceleration=3.5 m/s^2

and S is the distance cover by the car before it stops.

Now plugging the values in the third equation of motion

0=32^2-(2*3.5*S)

S=146.28 m

Therefore the car will cover a distance of 146 m before it stops.

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