# My attempt seems right? (Kinematics)

1. Oct 7, 2010

### IntegrateMe

A racing car traveling with constant acceleration increases its speed from 10m/s to 50m/s over
a distance of 60 m. How long does this take?

A. 2.0 s
B. 4.0 s
C. 5.0 s
D. 8.0 s
E. The time cannot be calculated since the speed is not constant

I did:

x - xo = 0.5 (Vo + V)(t)

Which is:

60 = 0.5 (60) (t)

Which is:

60 = 30t
t = 2

What am I doing wrong?

2. Oct 8, 2010

### Mindscrape

x=v0t+1/2*at^2

a= (v-v0)/t

combine and simplify
x=1/2(v+v0)t

t=(2x)/(v+v0)=2

3. Oct 8, 2010

### IntegrateMe

Are you sure? The answers came from my college textbook, but I suppose they could be wrong?

4. Oct 8, 2010

### Mindscrape

Well, a constant acceleration makes for a velocity with a linear slope, and the slope of that velocity is by definition the acceleration (it's not even the mean acceleration as long as acceleration is constant). The slope of the velocity is ∆v/∆t=40/t m/s^2 (since t starts at 0).

Plug this into the kinematic equation, and, of course, you get the same thing as the algebra.
60=10*t+20*t

you can also get a directly
(v^2-v0^2)/(2*x)=a

a=20m/s^2

if you plug this into the quadratic kinematic equation

0=-60+10t+10t^2

t=-3,2 (-3 is non-physical)

Either way, as you've posted the problem, the answer is most definitely 2s.