Math/Physics Problem

• Zeywery

  29 Oct 10, 10:21

  The distance between point A and B is 1000km.

  A car is travelling from point A to point B at a velocity of 14 km/h.

  The car accelerates about 2 km/h^2 every 1 hour.

  So what is the time taken for the car to travel from point A to point B?

   

• Forbiddensinner

  29 Oct 10, 11:15

  Originally posted by Zeywery:

  The distance between point A and B is 1000km.

  A car is travelling from point A to point B at a velocity of 14 km/h.

  The car accelerates about 2 km/h^2 every 1 hour.

  So what is the time taken for the car to travel from point A to point B?

   

  Lets use an unorthodox method. 

  The car is travelling like this:

  14 + (14+2) + (14+2+2) + (14+2+2) + ....

  Notice we have:

  Summation of 14 + 2r from zero to n.

  Rearranging, we will have:

  14(n+1) + 2(1+n)(n/2)

  This is equivalent to 1000, so we have:

  14n + 14 + n^2 + n = 1000

  n^2 + 15n - 986 = 0

  I will leave you to have fun using the quadratic formula.

  You should know which answer to reject and which to keep.

  The one remaining should be n = 1/2 (sqrt(4169)-15)

  Here is the tricky part.

  You will likely forget to add 1 to n to get the actual amount of hours.

  Notice that the summation is from zero to n.

  Hence, the time should be (n + 1) hours.

• ThunderFbolt

  30 Oct 10, 10:27

  or easier would be to just memorize the s = ut + 0.5 at^2...

• Zeywery

  31 Oct 10, 13:36

  Originally posted by Forbiddensinner:

  Lets use an unorthodox method. 

  The car is travelling like this:

  14 + (14+2) + (14+2+2) + (14+2+2) + ....

  Notice we have:

  Summation of 14 + 2r from zero to n.

  Rearranging, we will have:

  14(n+1) + 2(1+n)(n/2)

  This is equivalent to 1000, so we have:

  14n + 14 + n^2 + n = 1000

  n^2 + 15n - 986 = 0

  I will leave you to have fun using the quadratic formula.

  You should know which answer to reject and which to keep.

  The one remaining should be n = 1/2 (sqrt(4169)-15)

  Here is the tricky part.

  You will likely forget to add 1 to n to get the actual amount of hours.

  Notice that the summation is from zero to n.

  Hence, the time should be (n + 1) hours.

  How do you rearrange it to get that formula? Thanks

• wee_ws

  31 Oct 10, 14:23

  Hi,

    Recall two important results in sigma notation:

    1. sum k {r = a to r = b} is (b - a + 1)(k), where k is a constant.

    2. sum r {r = 1 to r = n} is (n/2)(1 + n).

    Thanks.

  Cheers,
  Wen Shih