Question

Car A is traveling at 18 m/s and car B at 25 m/s. Car A is 300m behind car B when the driver of Car A accelerates his car with an acceleration of 1.80m/s^2.
How long does it take car A to overtake car B?

Answer 1

Use the equation \[s= \frac{1}{2}at^2 +ut\]
Where u=initial velocity

Answer 2

s= distance; a= acceleration , and t =time

Answer 3

Rearrange the equation, then solve.

Answer 4

ok, is s like the intial distance?

Answer 5

s = total distance

Answer 6

In this case, the total distance that car A will travel is 300m, hence s=300m

Answer 7

so, i got 300t + 0.9t^2

Answer 8

wait nvm, its actually 0.9t^2 + u^2 - 300 ?

Answer 9

Close ...

check the u^2 ..

Answer 10

ohh. 0.9^2 + 18^t - 300

Answer 11

0.9t^2 rather

Answer 12

Yup. Good.

Answer 13

Make it equal to 0, then solve using the quadratic formula.

Answer 14

\[\frac{ -18\pm \sqrt{32.9} }{ 1.8 }\]

Answer 15

Check the number which is being square-rooted.

Answer 16

oh there shouldnt be a sq root over 32.9

Answer 17

It should be \[\sqrt{b^2-4ac}\]

Answer 18

\[\sqrt{18^{2}-4(0.9)(-300)}\]

Answer 19

Yep. Now put it into the quadratic formula

Answer 20

And remember, your answer should be positive, because you're trying to find the time, and time is always positive.

Answer 21

ok, so it should look like this?
\[\frac{ -18\pm37.5 }{ 1.8 }\]

Answer 22

and the positive answer is 10.8 secs?

Answer 23

Correct. =)

Answer 24

Thank you very much :P

Answer 25

You're welcome!