Car covered 40 m/s in 8.50 s while smoothly slowing down to 2.80 m/s. Find its original speed and then acceleration.

To find the original speed of the car, we can use the equation:

v_f = v_i + at

where:
v_f = final speed = 2.80 m/s
v_i = initial speed (what we are trying to find)
a = acceleration
t = time = 8.50 s

We can rearrange the equation to solve for v_i:

v_i = v_f - at
v_i = 2.80 m/s - a(8.50 s)

Now, we can use the information given to set up another equation:

40 m/s = v_i

Now we can substitute the expression for v_i into the second equation:

40 m/s = 2.80 m/s - a(8.50 s)

Now we can solve for the acceleration:

37.20 m/s = a(8.50 s)
a = 4.38 m/s^2

Therefore, the original speed of the car was 40 m/s and the acceleration was 4.38 m/s^2.

Step 1: Find the initial velocity (Vi)

Using the formula: Vf² = Vi² + 2 - a - d, where:

- Vf = final velocity (2.80 m/s)
- Vi = initial velocity (unknown)
- a = acceleration (unknown)
- d = distance (40 m)

Plugging in the values, we get:

(2.80 m/s)² = Vi² + 2 - a - 40 m

Vi² = 7.84 m² / s² - 80 - a m² / s²

Vi² = 7.84 m² / s² - 6400 - a (m²/s³) - (s²)

Step 2: Find the acceleration (a)

We don't have direct information about the acceleration, but we can use the time taken to complete the distance to find it.

Using the formula: Vf = Vi + a - t, where:

- Vf = final velocity (2.80 m/s)
- Vi = initial velocity (from Step 1)
- a = acceleration (unknown)
- t = time (8.50 s)

Plugging in the values, we get:

2.80 m/s = Vi + a - 8.50 s

a = (2.80 m/s - Vi) / 8.50 s

Step 3: Substitute the expression for acceleration into the equation from Step 1

Vi² = 7.84 m² / s² - 6400 - [(2.80 m/s - Vi) / 8.50 s]² - (s²)

This is a quadratic equation in Vi. Solving for Vi gives us:

Vi = 47.6 m/s (rounded to two significant figures)

Therefore, the original speed of the car was 47.6 m/s.

Substituting this value of Vi back into the equation from Step 2, we get:

a = -6.84 m/s² (rounded to two significant figures)

Therefore, the acceleration of the car was -6.84 m/s², indicating that it was decelerating.