Question

A tourist has planned a trip to cover the distance of 640 miles, driving at some constant speed. However, when he already covered a quarter of the distance, he took a rest for 1.2 hours. Then, in order to arrive at the destination on time, he increased the speed by 20 mph. How long, actually, the trip lasted?

Answer 1

Answer:

The trip lasted for a total of 8 hours.

Step-by-step explanation:

Distance planned = S = 640 miles

Constant speed = V

Thus the time to be taken would be = T = V/S

We have an equation 640 = VT ----- eq (a)

Time For First Quarter = T/4

speed = V

Distance = 640/4 = 160

After first quarter, there is a rest of 1.2 hours and to complete his trip on time, he increased the velocity by 20 mph.

So, the remaining distance = 640 - 160 = 480 miles.

Speed = V + 20 mph

Time remaining = [(T-T/4) - 1.2] = 3T/4 - 1.2 hours

We have an equation for remaining distance s = vt

=> 480 = (V+20)(3T/4 - 1.2) ----- eq (b)

using eq (a), we have V = 640/T. Putting it in eq (b), we have:

$480 = (\frac{640}{T} + 20)(3\frac{T}{4} - 1.2)\\480 = 480 - \frac{768}{T}  + 15T - 24\\=> 15T - \frac{768}{T} -24 = 0\\=> 15T^{2} - 24T - 768 = 0$

Solving the equation, we get T = 8 or T = -32/5(which is not possible.

So, the right answer is T = 8 hours