Ared car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis.,theredcarisatxr 220 m. if the red car has a constant velocity of 20 km/h, the cars pass each other at x 44.5 m, and if it has a constant velocity of 40 km/h, they pass each other at x 76.6 m. what are (a) the initial velocity and (b) the constant acceleration of the green car?

a) The initial velocity of the green car is -13 m/s

b) The acceleration of the green car is - 2.25 m/s²

Explanation:

The equation for the position of objects moving in a straight line with constant acceleration is as follows:

x = x0 + v0·t + 1/2·a·t²

where:

x = position

x0 = initial position

v0 = initial velocity

a = acceleration

t = time

If the velocity is constant, then a = 0 and x = x0 + v·t

a)The initial position of the red and green car is 0 m and 220 m respectively. We know that at 44.5 m the cars pass each other if the red car has a constant velocity of 20 km/h. So let´s find how much time it takes the cars to pass each other in this case:

The position of the red car is:

x = x0 + v·t

then:

0.0445 km = 0 km + 20 km/h · t

t = 0.0445 km/ 20 km/h = 8.0 s

We also know that if the red car has a velocity of 40 km/h, both cars pass each other at 76.6 m. So let´s find the time it takes the cars to reach that position using the equation for the red car:

0.0766 km = 0 km + 40 km/h · t

t = 0.0766 km / 40 km/h = 6.9 s

The position of the green car at t= 6.9 s and t = 8.0 s must be the same as the red car because both cars pass each other at those times.

Then, for the green car:

x = x0 + v0·t + 1/2·a·t²

0.0445 km = 0.220 km + v0 · 8.0 s + 1/2·a· (8.0 s)²

and

0.0766 km = 0.220 km + v0 · 6.9 s + 1/2·a· (6.9 s)²

Now we have a system of two equations with two unknowns.

Solving for "a" in the first equation

0.0445 km - 0.220 km - v0 · 8.0 s = 32 s²·a

(-0.176 km - v0 · 8.0 s) / 32 s² = a

Replacing a = (-0.176 km - v0 · 8.0 s) / 32 s² in the second equation and solving for v0:

0.0766 km = 0.220 km + v0 · 6.9 s + 1/2·((-0.176 km - v0 · 8.0 s)/32 s²)·(6.9 s)²

-0.143 km = v0 · 6.9 s - 0.74(0.176 km + v0 · 8.0 s)

-0.143 km = v0 · 6.9 s - 0.130 km - v0 · 5.9 s

-0.143 km + 0.130 km = v0 · 6.9 s - v0 · 5.9 s

-0.013 km = 1 s · v0

v0 = -13 m/s

b) The acceleration of the green car is:

a = (-0.176 km - v0 · 8.0 s) / 32 s²

a = (-0.176 km - (-0.013 km/s) · 8.0 s) / 32 s² = -2.25 m/s²

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.