1. ## Conversion Factors

A car is going 56km/h, it is accelerating at a rate of $\displaystyle {6m/s^2}$. How long would it take for it to obtain a speed of 90km/h?

For some reason i have some sort of brain block when it comes to conversion factors. Here is my attempt at an answer.

56km/h X $\displaystyle \frac{1000m}{1km}$ X $\displaystyle \frac{1hr}{3600s}$=15.56m/s

90km/h X $\displaystyle \frac{1000m}{1km}$ X $\displaystyle \frac{1hr}{3600s}$= 25m/s

25m/s-15.56m/s= 9.44m/s
$\displaystyle \frac{9.44m/s}{6m/s^2}$=1.57s

2. Originally Posted by mode
A car is going 56km/h, it is accelerating at a rate of $\displaystyle {6m/s^2}$. How long would it take for it to obtain a speed of 90km/h?

For some reason i have some sort of brain block when it comes to conversion factors. Here is my attempt at an answer.

56km/h X $\displaystyle \frac{1000m}{1km}$ X $\displaystyle \frac{1hr}{3600s}$=15.56m/s

90km/h X $\displaystyle \frac{1000m}{1km}$ X $\displaystyle \frac{1hr}{3600s}$= 25m/s

25m/s-15.56m/s= 9.44m/s
9.44/6m/s=1.57s
This is not a dimensional anaylsis problem since you are given both speeds in the same units.

Use Newton's Laws of Kinematics. You are given inital speed, acceleration (I assume it is constant) and you asked to find a final speed of 90.

v = 90
u = 56
a = 6
t = t

$\displaystyle v = u + at$

Solve for t

3. i guess the laws of kinematics would work for this question. But the focus of my lectures right now are conversion factors. let me get a better question up. give me a second

4. Originally Posted by e^(i*pi)
This is not a dimensional anaylsis problem since you are given both speeds in the same units.

Use Newton's Laws of Kinematics. You are given inital speed, acceleration (I assume it is constant) and you asked to find a final speed of 90.

v = 90
u = 56
a = 6
t = t

$\displaystyle v = u + at$

Solve for t
... velocity units have to be in m/s since acceleration was given in m/s^2.

the conversion still needs to be done ...

delta v = (90 - 56) = 34 km/hr = 34000/3600 = 9.44 m/s

1.57 s is correct.

5. How did I not see that? -facepalm-

It's time I went to bed I think.

Anyway, thanks for pointing it out

6. here is a better one that ive been having a lot of of problems with:
Concrete is sold by the cubic yard. What is the mass, in kilograms, of 8 cubic yards of concrete that is five times as dense as water? (1 m = 1.094 yd, and $\displaystyle {1 m^3}$ of water has a mass of 1,000 kg.

8yd^3= $\displaystyle \frac{1.094m}{1yd}$^3(fraction raised by power of 3)

= $\displaystyle {10.47m^3}$

10.47m^3 X 5000= 52,350kg
Im not sure where i went wrong, the answer btw is 30,550kg

7. 1.094 yd = 1 m ... you have it backwards

btw ... start a new problem w/ a new post.

8. whoa ok thanks a lot. i know for next time