The driver of a car traveling at 30 m/s needs to quickly reduce the car’s speed to 20m/s to round a curve in the highway. If the car’s deceleration has a magnitude of 6.0 m/s2, how far does it travel while reducing its speed?

The time required to decelerate by that amount is (30 - 20)/6.0 = 1.667 seconds

Multiply that by tha average speed, 25 m/s, for the distance traveled.

To find the distance traveled while reducing the car's speed, we can use the kinematic equation:

\(v^2 = u^2 + 2as\)

where
\(v\) is the final velocity (20 m/s),
\(u\) is the initial velocity (30 m/s),
\(a\) is the acceleration (deceleration in this case of 6.0 m/s²),
and \(s\) is the distance traveled.

Rearranging the equation, we have:

\(s = \frac{v^2 - u^2}{2a}\)

Now we substitute the known values:

\(s = \frac{(20^2) - (30^2)}{2 \times 6.0}\)

\(s = \frac{400 - 900}{12.0}\)

\(s = \frac{-500}{12.0}\)

\(s = -41.7 \, \text{m}\)

The negative sign indicates that the distance is in the opposite direction of motion. Therefore, the car travels 41.7 meters while reducing its speed.

To find the distance traveled by the car while reducing its speed, you can use the equation of motion:

v^2 = u^2 + 2as

Where:
v = final velocity (20 m/s)
u = initial velocity (30 m/s)
a = acceleration (deceleration in this case, which is -6.0 m/s^2)
s = distance

Rearranging the equation to solve for s, we have:

s = (v^2 - u^2) / (2a)

Now, substitute the given values into the equation:

s = (20^2 - 30^2) / (2 * (-6.0))

Simplifying further, you get:

s = (400 - 900) / (-12)

s = -500 / (-12)

s = 41.67 meters

Therefore, the car travels approximately 41.67 meters while reducing its speed.