Can you please help me with these five physics word problems? Thank you very much.

1. Let

• v₀ = initial velocity of the car

• a = 2.3 m/s², its acceleration

• x = 22.7 m, the distance covered by the car in 1.3 s

Since acceleration is constant, the distance covered after this time is

x = v₀t + 1/2 at ²

Solve for v₀ :

x = v₀t + 1/2 at ²

v₀t = x -  1/2 at ²

v₀ = x/t -  1/2 at

Plug in everything you know:

v₀ = (22.7 m)/(1.3 s) - 1/2 (2.3 m/s²) (1.3 s) ≈ 16 m/s

2. Recall that under constant acceleration,

v ² - v₀² = 2ax

where v is the velocity after some unknown time. This time,

• v₀ = 0 because the is stopped at the light

• a = 3.7 m/s²

• x = 47.0 m

Solve for v :

v ² - v₀² = 2ax

v ² = v₀² + 2ax

v = √(v₀² + 2ax)

Plug in everything you know:

v = √(0² + 2 (3.7 m/s²) (47.0 m)) ≈ 19 m/s

3. Use the same equation as in the previous problem, but this time you solve for a.

v ² - v₀² = 2ax

a = (v ² - v₀²)/(2x)

We're given

• v₀ = 23 m/s

• v = 0, because the boy wants to stop his bike

• x = 80 m

Then

a = (0² - (23 m/s)²)/(2 (80 m)) ≈ -3.3 m/s²

That is, the brakes have to apply a force that decelerates the bike with magnitude 3.3 m/s².

4. Same formula for acceleration as before,

a = (v ² - v₀²)/(2x)

with

• v = 17.3 m/s

• v₀ = 0

• x = 48.4 m

We end up with

a = ((17.3 m/s)² - 0²)/(2 (48.4 m)) ≈ 3.09 m/s²

5. By definition of average acceleration,

a (ave) = ∆v/∆t

That is, average acceleration is the ratio of change in velocity to change in time. Here we have

• ∆v = 0 - 38.0 m/s = -38.0 m/s, the difference between the final and initial velocities

• ∆t = 3.9 s, the time it takes to stop

Then

a (ave) = (-38.0 m/s)/(3.9 s) ≈ -9.7 m/s²

what is the actual question though?