Speed, Time And Distance Quiz-1

Speed of yatch downstream = \frac {Distance covered}{Time taken}

= \frac {45}{2} = 12.9 km/h

Now, speed of boat upstream = \frac {Distance covered}{Time taken}

= \frac{45}{7.1} = 6.4 km/h

Now, speed of boat = Speed downstream+Speed upstream

= 12.9 + 6.4 = 19.3 km/h

Let the speed of goods train be x km/hr.

Then, (67+x) \times \frac{5}{18} = \frac{293.5}{11}

⇒ x= 29.05 km/hr

Let the speed of the stream be x km/hr. Then,

Speed downstream = (21 + x) km/hr,

Speed upstream = (21 - x) km/hr.

\frac{45}{21+ x} + \frac{45}{21 - x} = 4\tfrac{1}{2}

\frac{2025}{441 - x^{2}} = \frac{9}{2}

9x^{2} = 181

x = 4.48 km/hr.

Let the speed of the cyclist be x km/h and wind be y km/h

\frac{1}{x+y} = \frac{3}{60}

x+y=20..........(i)

And

\frac{1}{x-y} = \frac{4}{60}

x-y = 15.........(ii)

On solving both the equations we get:

x = \frac{35}{2} , y = \frac{5}{2}

Now, time taken by cyclist without wind: \frac{2}{35} \times 60 = \frac{24}{7} min

Let the car travel a distance of 2d km, total.

The first half of the distance takes \frac{d km}{30 kmph} =\frac{d}{30} hr.

The second half of the distance takes \frac{d km}{50 kmph} =\frac{d}{50} hr.

So the total time taken to cover 2d km = \frac{d}{30} + \frac{d}{50} = \frac{80d}{150} hr.

The average speed of the car = \frac{Total distance}{total time} = \frac{2d}{\frac{80d}{150}}

= \frac{2d \times 150}{80d}

= 37.5 kmph.

Let the speed of the current be x km/hr

Thus upward speed =(6+x) km/hr and downward speed =(6−x) km/hr

Let distance traveled be y, then

\frac{y}{6−x} = \frac{3y}{6+x}

⇒ x= 6 km/hr

Given speed = 71 km/hr = 71 \times \frac{18}{5} = 255.6 m/s

Let the length of the bridge = x mts

Given time taken to cover the distance of (183 + x)mts is 30 sec.

We know speed = \frac{distance}{time} m/s

255.6 = 183 + 30x
x = 2.42 mts.

Let the length of the train be x

\frac{x+142}{38} = \frac{x}{23}

=> 23(x + 142) = 38x

=> 23x + 3266 = 38x

=> 3266 = 15x

=> x = 217.73 m.

Let the distance of train be any value ( if you don't want to solve by taking x. )

Let length of each train = L.C.M. of the the time given 29 and 19 (for simplicity) = 551 meter.

Passing a pole means, crossing its own length.

So, for train 1.

Distance covered in passing a pole = length of train 1 =551 meter.

Time taken = 29 sec.

So, Speed of train 1 = \frac{551}{29} = 19 m/s

For train 2:

Speed of train 2 = \frac{551}{19} = 29 m/s

( These data can be assumed in solving relatively. If x would have been taken in place of 551, the process would have been the same. If confused, think x=551. )

Opposite direction movement, time taken to cross:

In opposite direction, the relative speed is added up.

So, relative speed = 19 + 29 = 48 m/s

Relative Distance covered for both trains to cross each other = sum of length of both trains = 551 + 551 = 1102 m

So, time taken for both trains to cross each other= \frac{distance}{relative speed} = \frac{1102}{48} sec = 22.9 sec.

Since the train and man are moving in opposite directions,

The relative speed = 75 - (-8.8) kph = 75 + 8.8 = 83.8 kph

83.8 km/h in m/s = 83.8 \times \frac{5}{18} = 23.27m/s

Total distance to be covered = 135 m (length of the train)

So, time = \frac{distance}{speed} = \frac{135}{23.27} =  5.80 s.

So, the required time is 5.80 seconds.