the time required for a car to accelerate from rest to 8.3m/s at 0.75m/s^2 (meters per second cubed) is....?

divide 8.3/0.75 to get 11.1

its easy u idiot

To find the time required for the car to accelerate from rest to 8.3 m/s at an acceleration of 0.75 m/s^2, we can use the following kinematic equation:

v = u + at

where:
v = final velocity (8.3 m/s)
u = initial velocity (0 m/s, as the car starts from rest)
a = acceleration (0.75 m/s^2)
t = time

Substituting the known values into the equation, we have:

8.3 m/s = 0 m/s + (0.75 m/s^2) * t

Simplifying, we have:

8.3 m/s = 0.75 m/s^2 * t

To isolate t, divide both sides of the equation by 0.75 m/s^2:

t = 8.3 m/s / 0.75 m/s^2

Simplifying further, we get:

t = 11.067 seconds (rounded to 3 decimal places)

Therefore, the time required for the car to accelerate from rest to 8.3 m/s at an acceleration of 0.75 m/s^2 is approximately 11.067 seconds.

To find the time required for the car to accelerate from rest to a certain velocity, we can use the formula:

$v = u + at$

Where:
- $v$ is the final velocity (8.3 m/s in this case)
- $u$ is the initial velocity (0 m/s since the car starts from rest)
- $a$ is the acceleration (0.75 m/s^2 in this case)
- $t$ is the time taken

We need to rearrange the formula to solve for $t$:

$t = \frac{v - u}{a}$

Now, let's substitute the given values:

$t = \frac{8.3 \, \text{m/s} - 0 \, \text{m/s}}{0.75 \, \text{m/s}^2}$

Simplifying the equation:

$t = \frac{8.3 \, \text{m/s}}{0.75 \, \text{m/s}^2}$

$t ≈ 11.07 \, \text{s}$

Therefore, the time required for the car to accelerate from rest to 8.3 m/s at an acceleration of 0.75 m/s^2 is approximately 11.07 seconds.