Question

A car traveling 35.0 m/s starts to accelerate at 3.00 m/s2. How far does this car travel after 4.00 seconds?

Answer 1

To find the distance the car travels after 4.00 seconds of acceleration, you can use the following equation of motion:

`\[d = v_i t + (1)/(2) a t^2\]`

Where:

-
`\(d\)` is the distance traveled,

-
`\(v_i\)` is the initial velocity (35.0 m/s),

-
`\(a\)` is the acceleration (3.00 m/s²), and

-
`\(t\)` is the time (4.00 seconds).

Now, plug in the values:

`\[d = (35.0 m/s)(4.00 s) + (1)/(2)(3.00 m/s²)(4.00 s)^2\]`

`\[d = 140.0 m + (1)/(2)(3.00 m/s²)(16.0 s^2)\]`

`\[d = 140.0 m + 24.0 m\]`

`\[d = 164.0 meters\]`

So, the car travels a distance of 164.0 meters after 4.00 seconds of acceleration.