# T & J travel from Bedk to Jrong, distance of 34km. T is moving at 1/2 km h faster than J and it took T half an hour earlier to Jrong. Speed of T?

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### 1 Answer

The distance between Bedk and Jrong is 34 km.

Let the speed of J be S k/h. Then the speed of T is (S+1/2) k/h

Now, Let the time needed for J be X hours. Then, the time needed for T is (X- 1/2) hours.

Now we know that:

Distance = speed * Time

==> J's distance = 34 = S*x

==> T's distance = 34= (S+1/2)* (X-1/2)

==> s = 34/x

==> s+ 1/2 = 34/(x-1/2)

==> 34/x + 1/2 = 34/(x-1/2)

==> (68 + x)/2x = 34/(x-1/2)

==> (68 + x)/2x = 68/(2x-1)

==> (2x-1)(x+68) = 2x*68

==> (2x^2 + 136x - x - 68 = 136x

==> 2x^2 - x - 68 = 0

Now we will use the quadratic formula to find x.

==> x1= (1 + sqrt(1+4*2*68)/ 2*2 = (1+ 23.35)/ 4 = 24.35/4

==> x1= 6.08

Then the time needed for J is x = 6.08 hours

==> S= 34/ 6.08= 5.59 km per hour

**==> Then the speed needed for T is S+ 1/2 = 5.59 + 0.5 = 6.08 km per hour.**