Challenging Speed, Time, Distance Questions (Hints & Answers)

Let the distance between stations = d Km.

Speed from A to B = 200 Km/h

Speed from B to A = 200 Km/h + 100 Km/h = 300 Km/h

Can you find the time taken from A to B and B to A? What about total time?

Ans.  240 Km/h

Can you find the relation between the original speed and time till bus hits the road block?

Ans.  80Km/h

Still water neither propels nor resists the boat in any direction. However, traveling downstream, the boat is going with the flow of water. It propels the boat with it. Its speed gets added to the speed of the boat.

Let the speed of water be x Km/h.

Therefore downstream boat speed = 5 + x Km/h.

Can you now find the time taken to travel downstream?

Ans.  1 Km/h

In how much time does the drone meet Sofia? What is the distance traveled by the drone during this time?

Ans.  13.5 Km

Direction Of Boat:

If the boat travels on the straight line AB, by the time it will reach the opposite side, the river will carry it forward to some point D. Therefore, it has to travel on the line AC, to reach point B, directly opposite to point A.

Let the time taken by the boat to complete the trip = t

Can you find distances AC and CB in terms of t?

Ans.  40 seconds

Let the distance between stations = d Km.

Speed from A to B = 200 Km/h

Speed from B to A = 200 Km/h + 100 Km/h = 300 Km/h

From the speed formula:

Time = Distance/Speed

Time from A to B = t_AB = d/200

Time from B to A =t_BA = d/300

Total Time = t_AB + t_BA = d/200 + d/300

$=\frac{300d+200d}{200\times 300}=\frac{500d}{200\times 300}$ hours

Let us calculate the average speed now.

Total distance = AB + BA = d + d = 2d

AverageSpeed = TotalDistance/TotalTime

$=\frac{2d}{\frac{500d}{200\times 300}}$

$=\frac{2d\times 200\times 300}{500d}$

$=\frac{400\times 3}{5}=240$ Km/h

Average Speed = 240 Km/h

The bus stops at point C for 1/2 hour due to blockage.

Let the initial speed be x Km/h

Time = Distance/Speed

Time to travel the distance AC = t_AC = 300/x hours

Let’s first calculate CB

60% of AB = 300 Km

AB × 0.60 = 300

AB = 300/0.60 = 500 Km

CB = Remaining distance = 500 − 300 = 200 Km

Speed of bus after point C = (x + 20) Km / h

Time to travel the distance CB = t_CB = 200/(x + 20) hours

Had there been no blockage, the bus would have traveled at xKm/h. Its travel time would have been:

t_usual = AB/x = 500/x hours

The bus reached town B on time, despite blockage, therefore:

Usual time = Time to travel the distance AC + Delay + Time to travel the distance CB

$t_{usual}=t_{AC}+\frac{1}{2}+t_{CB}$

$\frac{500}{x}=\frac{300}{x}+\frac{1}{2}+\frac{200}{x+20}$

Multiplying both sides by 2x(x + 20):

500 × 2(x + 20) = 300 × 2(x + 20) + x(x + 20) + 200 × 2x

1000x + 20000 = 600x + 12000 + x² + 20x + 400x

1000x + 20000 = x² + 1020x + 12000

0 = x² + 1020x − 1000x + 12000 − 20000

x² + 20x − 8000 = 0

The speed is calculated by solving the above quadratic equation. Factorizing the quadratic expression we get:

(x + 100)(x − 80) = 0

x = − 100, or x = 80.

Speed is not a negative quantity, so x = 80Km/h.

Still water neither propels nor resists the boat in any direction. However, traveling downstream, the boat is going with the flow of water. It propels the boat with it. Its speed gets added to the speed of the boat.

Let the speed of water be x Km/h.

Therefore downstream boat speed = 5 + x Km/h.

Downstream Distance = 18 Km

Downstream Speed = (5 + x) Km/h

Downstream Time = $t_{down}=\frac{DownstreamDistance}{DownstreamSpeed}$

$t_{down}=\frac{18}{5+x}$ hours

While traveling upstream, the boat is going against the flow of water. So water carries it in the opposite direction. The reduction in its speed is equal to the speed of the water. Therefore the speed of the boat upstream is 5 − x Km/h.

Upstream Distance = 16 Km

Upstream Speed = (5 − x) Km/h

Upstream Time = $t_{up}=\frac{UpstreamDistance}{UpstreamSpeed}$

$t_{up}=\frac{16}{5-x}$ hours

Total time = t_down + t_up = 7 Hours

$\frac{18}{5+x}+\frac{16}{5-x}=7$

$\frac{18(5-x)+16(5+x)}{(5+x)(5-x)}=7$

90 − 18x + 80 + 16x = 7(5 + x)(5 − x)

−2x + 170 = 7(25 − x²)

−2x + 170 = 175 − 7x²

7x² − 2x − 5 = 0

(7x + 5)(x − 1) = 0

(7x + 5) = 0 Or (x − 1) = 0 Therefore $x=\frac{-5}{7}$ Or x = 1

As the speed of water cannot be negative, x = 1 is the only acceptable solution.

Speed of Water = 1 Km/h

Let’s say the Maya starts from A and Sofia from B

When the drone meets Sofia at point C, Maya reaches point D

Drone flies back to meet the Maya at point F

Distance flown by drone = AC + CF

Speed of Drone = $s_{drone}=15$ Km/hr

Speed of Maya/Sofia = $s_{walk}=5$ Km/hr

Let’s say drone and Sofia travel toward each other for time $t_{AC}$ to meet at point C.

Distance traveled by drone = AC = $s_{drone}\times t_{AC}=15\times t_{AC}$

At the same time, Sofia reaches point C.

Distance traveled by Sofia = CB = $s_{walk}\times t_{AC}=5\times t_{AC}$

Total distance traveled by both = AC + CB = AB = 12 Km

$15t_{AC}+5t_{AC}=12$

$20t_{AC}=12$

$t_{AC}=\frac{20}{12}=\frac{3}{5}$

Therefore AC = $15\times t_{AC}=15\times \frac{3}{5}=9$ Km

By the time drone reaches point C, the Maya reaches point D.

AD = $s_{walk}\times t_{AC}$

AD = $5\times \frac{3}{5}=3$ Km

The distance remaining between drone and Maya:

$=DC=AC-AD=9-3=6$ Km

Both meet at point F, after time $t_F$.

$DC=DF+FC=6Km$ Equation 1

Distance traveled by the drone in time $t_F$:

CF = FC = $s_{drone}\times t_F=15\times t_F=15t_F$ Equation 2

Distance traveled by Maya in time $t_F$:

DF = $s_{walk}\times t_F=5\times t_F=5t_F$ Equation 3

Putting values of FC and DF (from equation 2 and 3) in Equation 1:

$6=5t_F+15t_F=20t_F$

$t_F=\frac{6}{20}$

Using equation 2, CF = $15t_F=15\times \frac{6}{20}=4.5$ Km

Let us finally calculate the total distance traveled by the drone:

= AC + CF = 9 + 4.5 = 13.5 Km

Direction Of Boat:

If the boat travels on the straight line AB, by the time it will reach the opposite side, the river will carry it forward to some point D. Therefore, it has to travel on the line AC, to reach point B, directly opposite to point A.

Let the time taken by the boat to complete the trip = t

Distance Calculation:

The distance river carries the boat during this time = CB = SpeedOfRiver × t

CB = 12t

The distance traveled by boat = AC = SpeedOfBoat × TimeTaken

AC = 37t

As unit of speed is m/s we convert the width of river (= AB) to meters.

AB = 1.4 Km = 1400 m

From the Pythagoras theorem we have:

AC² = CB² + AB² = (12t)² + (1400)²

(37t)² = (12t)² + (1400)²

1359t² − 144t² = (1400)²

1225t² = (1400)²

$t=\frac{1400}{\sqrt{1225}}=40$ seconds

The students familiar with vectors can try to solve the problem using the velocities of boat and river.