a car is speeding up and has an instantaneous velocity of 6.0 m/s in the x direction when a stopwatch reads 10.0 s. it has a constant acceleration of 1.5 m/s^2 in the x direction until the stopwatch 12.0 s. how far does it travel during this time interval?

call t=0 when the stopwatch reads 10. Then for the next two seconds, starting with initial velocity of 6,

s(t) = 6t + .75t^2
s(2) = 6*2 + .75*4 = 15.0m

To find the distance traveled by the car during the given time interval, we can use the equations of motion.

First, let's find the final velocity of the car. We know that it has an initial velocity of 6.0 m/s and a constant acceleration of 1.5 m/s^2. The formula to calculate the final velocity (v) with initial velocity (u), constant acceleration (a), and time (t) is:

v = u + at

Plugging in the values, we have:
v = 6.0 m/s + (1.5 m/s^2)(12.0 s)
v = 6.0 m/s + 18.0 m/s
v = 24.0 m/s

Now, we can find the distance traveled by the car using the formula:
s = ut + (1/2)at^2

Since the car has a constant acceleration, we can use the average velocity (v_avg) during the time interval to simplify the equation. The average velocity is given by the formula:
v_avg = (u + v) / 2

Plugging in the values, we have:
v_avg = (6.0 m/s + 24.0 m/s) / 2
v_avg = 30.0 m/s / 2
v_avg = 15.0 m/s

Now, we can use the average velocity to calculate the distance traveled:
s = v_avg * t

Plugging in the values, we have:
s = 15.0 m/s * 12.0 s
s = 180.0 m

Therefore, the car travels a distance of 180.0 meters during the given time interval.