A car accelerates from 10.0 m/s to 30.0 m/s  at a rate of 3.00 m/s2.How far does the car travel while accelerating?Express your answer in meters.

To solve this problem, we can use the equation of motion that relates final velocity, initial velocity, acceleration, and distance. The equation is:

v^2 = u^2 + 2as

where: v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.

We can rearrange the equation to solve for s:

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

Substituting the given values:

s = ((30.0 m/s)^2 - (10.0 m/s)^2) / (2 * 3.00 m/s^2) s = (900 m^2/s^2 - 100 m^2/s^2) / 6.00 m/s^2 s = 800 m^2/s^2 / 6.00 m/s^2 s = 133.33 m

So, the car travels approximately 133.33 meters while accelerating.

To solve this problem, we can use the equation of motion that relates final velocity, initial velocity, acceleration, and distance traveled. This equation is:

v^2 = u^2 + 2as

where: v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.

We can rearrange this equation to solve for s:

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

Substituting the given values:

s = ((30.0 m/s)^2 - (10.0 m/s)^2) / (2 * 3.00 m/s^2) s = (900 m^2/s^2 - 100 m^2/s^2) / 6.00 m/s^2 s = 800 m^2/s^2 / 6.00 m/s^2 s = 133.33 m

So, the car travels approximately 133.33 meters while accelerating.