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

Answer:

168 m

Explanation:

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

$s= Vi*t+(\frac{1}{2}Vft^{2} )$

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

$a=\frac{Vf-Vi}{t}$

We clear time and are left with:

$t=\frac{Vf-Vi}{a}$

Now we insert the values we know:

$t=\frac{Vf-Vi}{a}$

$t=\frac{18-6}{3}$

$t=\frac{12}{3}$

$t=4$

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

$s= Vi*t+(\frac{1}{2}Vft^{2} )$

$s= 6*4+(\frac{1}{2}(18)4^{2} )$

$s= 24+(\frac{1}{2}(18)16 )$

$s= 24+(144)$

$s= 168m$

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

Answer 2

 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