# Related to Constant acceleration !

1. Apr 26, 2012

### HelpMeHelpYou

1. The problem statement, all variables and given/known data

Suppose a child driving a go-cart is traveling 4m/s when she crosses a line 4m from her starting point. She continues with a steady acceleration of 0.4m/s^2 , until she crosses a mark 40m from the starting point. How long does it take for her to go from the 4m mark to the 40m mark?

what i did is this :

a = 0.4m/s^2
Vi = 4m/s
d = 40m - 4m = 36m

i used this formula :

Vf = Vi + a*t
0 = 4m/s + (0.4m/s^2)t
-0.4m/s^2 = 4m/s
t = -10s

and this :

d = Vi+Vt/2 * t
36m = (4m/s + 0)/2 * t
36m = 2m/s t

divide both side by 2m/s

t = 18s

but the true answer is 6.7s

anyone correct this ?

2. Apr 26, 2012

### sambristol

the direct way to solve it, without having to calculate the final velocity, is to use
d = V1 t + a t2 /2
put in the values d= 36, a=0.4 and solve the quadratic equation in t using the normal formula. Its obviously the positive value of t you take.

Hope this helps

Regards

Sam

Last edited: Apr 26, 2012
3. Apr 26, 2012

### HelpMeHelpYou

how about the velocity ? Vi = 0 or 4m/s ?

4. Apr 26, 2012

### sambristol

4 m/s

5. Apr 26, 2012

### andrien

36=4t+0.2t^2

6. Apr 26, 2012

### HelpMeHelpYou

just a confirmation ..

correct this if i'm wrong ..

d = 36
a = 0.4

i used :

d = V1 t + a t2 /2
subst:

36 = 0 t + 0.4 t2 /2
divide both sides by 0.4

90 = t2 /2
divide both sides by 1/2

45 = t2
square root both sides to remain t :

t = 3â5 or 6.7

is it correct usage of algebra ?

7. Apr 26, 2012

### sambristol

try 36 = 4t+0.2t2

0.2 t2 +4t -36 =0

t = (-4 + â(16+4*0.2*36))/(2*0.2)

tâ6.7 seconds

Last edited: Apr 26, 2012
8. Apr 26, 2012

### HelpMeHelpYou

whow ! again .. i should study back again .. so thanks Sam ..

gratitude !