SOLUTION: A delivery truck follows a regular route that is 270 km. One day the driver begins the route half hour late. In order to finish on time, she drives the truck 6km/hr faster than usu

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Question 1171220: A delivery truck follows a regular route that is 270 km. One day the driver begins the route half hour late. In order to finish on time, she drives the truck 6km/hr faster than usual. What is the truck's usual speed
Found 4 solutions by Boreal, ikleyn, josgarithmetic, MathTherapy:
Answer by Boreal(15235)   (Show Source):
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d=v*t
270=vt
and
270=(v+6)*(t-0.5)
therefore, vt=vt-0.5v+6t-3
and 0.5v=6t-3 so v=12t-6
therefore, 12t^2-6t-270=0 substituting
2t^2-t-45=0
(2t+9)(t-5)=0
t=5 hours
so normally the speed is 54 km/h
at 60 km/h, 6 km/h faster, the trip will take 4 1/2 hours.
54 km/h

Answer by ikleyn(52803)   (Show Source):
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.

The time equation is - = . Each term in the left side is the traveled time. in the right side represents half an hour time difference. Reduce to quadratic equation and solve it using quadratic formula or factoring. ANSWER. 54 km/h and 60 km/h.

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Answer by josgarithmetic(39620)   (Show Source):
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SPEED TIME DISTANCE NORMAL r 270/r 270 LATE r+6 270/(r+6) 270 DIFFERENCE 1/2

-------solve for r.

Answer by MathTherapy(10552)   (Show Source):
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A delivery truck follows a regular route that is 270 km. One day the driver begins the route half hour late. In order to finish on time, she drives the truck 6km/hr faster than usual. What is the truck's usual speed

Let truck's normal speed be S
Then usual time =
We then get the following TIME equation:
270(2)(S + 6) = 270(2S) + S(S + 6) ------ Multiplying by LCD, 2S(S + 6)

(S - 54)(S + 60) = 0
Normal speed of truck, or OR S = - 60 (ignore)