Application of integration

[dbag]

Problem statement:
Cars A and B are going in the same direction and are are next to each other at t=0 at which point car A has a velocity of 0 and car B has a velocity of 10. Car a has an acceleration of 2 and car B has an acceleration of 0,6t. Calculate at what value of t(time) will they be next to each other again.

Solution so far:
Step 1.
Car A a(t)=2 --> v(t)=2t+c --> c=v0 and v0 = 0 because v@ t=0 is 0
Car B a(t)=0,6t --> v(t)=0.3t^2+c --Z c=v0 and v0 is 10 because v@t0=10

Step 2.
Car A s(t)=t^2+c --> c=s0 and s0=0 and thats an assumption on my part
Car B s(t)= 0,9t^3+10t+c and c is also 0 for the reason above

Step 3.
In order to figure out the time at which these 2 cars meet again i just made s(t)a=s(t)b, so
t^2=0,9t^3+10t --> 0,9t^3-t^2+10t=0
and the solutions would follow: t=0, t= 3.286, t=-3.286
Am i missing anything? is it correct?
Thanks in advance

 

[Deleted member 4993]

Problem statement:
Cars A and B are going in the same direction and are are next to each other at t=0 at which point car A has a velocity of 0 and car B has a velocity of 10. Car a has an acceleration of 2 and car B has an acceleration of 0,6t. Calculate at what value of t(time) will they be next to each other again.

Solution so far:
Step 1.
Car A a(t)=2 --> v(t)=2t+c --> c=v0 and v0 = 0 because v@ t=0 is 0
Car B a(t)=0,6t --> v(t)=0.3t^2+c --Z c=v0 and v0 is 10 because v@t0=10

Step 2.
Car A s(t)=t^2+c --> c=s0 and s0=0 and thats an assumption on my part
Car B s(t)= 0,9t^3+10t+c and c is also 0 for the reason above

Step 3.
In order to figure out the time at which these 2 cars meet again i just made s(t)a=s(t)b, so
t^2=0,9t^3+10t --> 0,9t^3-t^2+10t=0
and the solutions would follow: t=0, t= 3.286, t=-3.286
Am i missing anything? is it correct?
Thanks in advance

Please check your result of integration:

\(\displaystyle \displaystyle{\int (0.3*t^2 + 10) dt}\)..................................edited

= \(\displaystyle \displaystyle{\frac{0.3}{3}t^3 +{10}t + C}\)..................................edited

Otherwise your method is correct..... so continue.....

 

[dbag]

Please check your result of integration:

\(\displaystyle \displaystyle{\int (0.3*t^2 + 10*t) dt}\)

= \(\displaystyle \displaystyle{\frac{0.3}{3}t^3 + \frac{10}{2}t^2 + C}\)

Otherwise your method is correct..... so continue.....

allrighty. instead of 0,9t³ it is 0,1t³ but that second part is wrong.
for car B the function for velocity v(t)=0,3t²+10, then s(t)=0,1t³+10t+c, and not 10/5t²