Question

13. A car accelerates from 6.0 m/s to 18 m/s at a rate of 3.0 m/s^2 How far does it travel while accelerating?

Answer 1

Final velocity = Initial velocity – 9.8 * t
Final velocity = 80.5 – 9.8 * 0.0124 = 80.378 m/s
OR
Final velocity = 80.5 – 9.8 * 16.416 = 80.378 m/s

A. -32 m/s
B. -97 m/s
C. -64 m/s
D. -81 m/s
E. -48 m/s

Answer 2

168 m

Step-by-step explanation:

First of all we need to rememeber the formula to calculate the distance with the acceleration:

`s= Vi*t+((1)/(2)Vft^(2)  )`

Now as we do not have time, we have to use first the formula for acceleration but cleared for time:

`a=(Vf-Vi)/(t)`

We clear time and are left with:

`t=(Vf-Vi)/(a)`

Now we insert the values we know:

`t=(Vf-Vi)/(a)`

`t=(18-6)/(3)`

`t=(12)/(3)`

`t=4`

Now that we have time we just have to insert the values into the first formula:

`s= Vi*t+((1)/(2)Vft^(2)  )`

`s= 6*4+((1)/(2)(18)4^(2)  )`

`s= 24+((1)/(2)(18)16  )`

`s= 24+(144)`

`s= 168m`

And the distance that he traveled while accelerating is 168m.