A driver slows down her car from 32.7 km/hr at a constant rate of 0.63 m/s2 just by taking her foot of the accelerator (the gas pedal). Calculate the distance the car travels during the sixth second (between t=5 and t=6 seconds)that she is coasting.

She moves 5.616 meters in that second.

Explanation:

If we define t = 0s as the moment when she starts decelerating we can write the function of acceleration as:

a(t) = -(0.63 m/s^2)

where the negative sign is because she is slowing down.

The velocity equation can be found if we integrate over time:

v(t) = -(0.63m/s^2)*t + v0

Where v0 is the constant of integration, that represents the initial velocity, in this case is:

v0 = 32.7 km/h

Now, because the acceleration is in m/s^2, we should write this velocity in m/s.

in one km we have 1000 meters, and in one hour we have 3600 seconds, then we have that:

32.7 km/h = 32.7 *(1000/3600) m/s = 9.08 m/s

Then the velocity equation becomes:

v(t) = -(0.63m/s^2)*t  + 9.08 m/s

And for the position equation, we should integrate again to get:

p(t) = -(1/2)*(0.63m/s^2)*t^2 + (9.08m/s)*t + p0

Where p0 is the initial position.

For this problem, we want to find the distance that she moved between t = 5s and t = 6s, and that can be calculated as:

D = p(6s) - p(5s)

D = -(1/2)*(0.63m/s^2)*(6s)^2 + (9.08m/s)*6s + p0 +(1/2)*(0.63m/s^2)*(5s)^2 - (9.08m/s)*(5s) - p0

D = -(1/2)*(0.63m/s^2)*((6s)^2 - (5s)^2) + (9.08m/s)*(6s - 5s)

D = 5.615 m

She moves 5.616 meters in that second.

object 2 because all the other ones are wrong

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.