Tag: time - calcuation of distance

Questions Related to time - calcuation of distance

A man rows to a place 48 km distant and comes back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. The rate of the stream is ............

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. 0.5 km/hr

  2. 1 km/hr

  3. 3.5 km/hr

  4. 1.8 km/hr

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Correct Option: B
Explanation:

Suppose he move 4 Km downstream in x hours.Then 
Speed downstream=$\frac{4}{n}$Km\hr
Speed upstream=$\frac{3}{n}$Km\hr
$\therefore \frac{48}{$\frac{4}{n}$}+\frac{48}{\frac{3}{x}}=14$ 
$\frac{48x}{4}+\frac{48x}{3}=14$
Or x=$\frac{336}{168}=2$
So spead downstream=$\frac{4}{\frac{1}{2}}=8$Km\hr
Or speed upstream =$\frac{3}{\frac{1}{2}}=6$Km\hr
So rate of stream =$\frac{1}{2}(8-6)=\frac{2}{2}=1$Km\hr

Fighter planes  $X$  and  $Y$  are moving towards a target $ 'O'$ along two perpendicular paths, with equal speeds.  $X$  starts from a point at a distance of  $19\mathrm { km }$  from  $^ { \prime } O ^ { \prime }$  and  $Y$  starts from a point at a distance of   $12\mathrm { km }$  from  $ \mathrm '{ O } '.$  After  $1$  minute, it was found that they were  $13\mathrm { km }$  away from each other. What is the speed at which they are travelling. given that they start simultaneously?

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. $35 \mathrm { km } / \mathrm { min }$

  2. $28 \mathrm { km } / \mathrm { min }$

  3. $7 \mathrm { km } / \mathrm { min }$

  4. $21 \mathrm { km } / \mathrm { min }$

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Correct Option: B

Standing on a platform Abdul told Nagma that Aligarh was more than ten kilometers but less than fifteen kilometers from there Nagma knew that it was more than twelve but less than fourteen kilometers from there If both of them were correct which of the following could be the distance of Aligarh from the platform?

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. 13 km

  2. 12 km

  3. 11 km

  4. 14 km

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Correct Option: A
Explanation:

According to Abdul Aligarh is more than 10km and less than 15 kms
According to Nagma, Aligarh is more than 12 km but less than 14 kms
If both are correct, then the only common number between them is 13 km
Answer is Option A

The distance-time relationship of a moving body is given by $y=F(x)$ then the acceleration of the body is the:

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. Gradient of the velocity/time graph

  2. Gradient of the distance/time graph

  3. Gradient of the acceleration/time graph

  4. Gradient of the velocity/distance graph

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Correct Option: A
Explanation:

By definition,


Gradient of a distance time graph gives VELOCITY.

Gradient of a velocity-time graph gives ACCELARATION. 

A car travels 120 km from A to B at 30 km per hour but returns the same distance at 40 km per hour. The average speed for the round trip is closest to: 

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. 33 km/hr

  2. 34 km/hr

  3. 35 km/hr

  4. 36 km /hr

  5. 37 km/hr

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Correct Option: B
Explanation:

If a car travels a distance d at rate $r _1$  and returns the same distance at rate $r _2$ , then 
Average speed = $\dfrac{total distance}{total time} = \dfrac{2d}{d/r _1+ d/r _2}= \dfrac{2r _1r _2}{r _1+r _2}$; 
$\therefore  x = \dfrac{2.30.40}{70} = \dfrac{240}{7} ~ 34 km/hr$.

A certain sum of money at simple interest amounts to $Rs. 1012$ in $2\dfrac {1}{2}$ years and to $Rs. 1067.20$ in $4$ years. The rate of interest per annum is

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. $2.5$%

  2. $3$%

  3. $4$%

  4. $5$%

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Correct Option: C
Explanation:

Let the principal be $P$ and rate of interest be $r$%.
According to question,
$1012 = P + \dfrac {P\times r \times 5}{100\times 2} .... (1)$
Interest in $\dfrac {3}{2} years = 1067.20 - 1012 = Rs. 55.20$
$P = \dfrac {I\times 100}{R\times T} = \dfrac {55.20\times 100}{3} = \dfrac {3680}{r}$
Putting values in equation $(1)$,
$1012 = \dfrac {3680}{r} + \dfrac {3680\times r\times 5}{r\times 100\times 2}$
$1012 = \dfrac {3680}{r} + 92$
$\dfrac {3680}{r} = 1012 - 92 = 920$
$r = \dfrac {3680}{920} = 4$% per annum.
Hence, the rate of interest is $4$% per annum.

Two bodies $A$ and $B$ are projected vertically up from the ground simultaneously. They spend $6$ seconds and $9$ seconds in air respectively. Raio of the maximum heights reached by them is

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. $2:3$

  2. $6:9$

  3. $12:27$

  4. $4:9$

Show answer
Correct Option: D
Explanation:
Let $A$ and $B$ have initial velocity as it $\mu _{2}$
As from Newtons first law of motion, $\upsilon = \mu t$ at
$\Rightarrow 0 = \mu _{1}- g t _{1} \Rightarrow t _{1} \dfrac{\mu _{1}}{g} (\upsilon = 0$ at top)
Similarly $t _{2}= \dfrac{\mu _{2}}{g}$
Now, from third law, $S= \dfrac{\upsilon^{2} - \mu^{2} }{2a}$
$\Rightarrow h ,  \dfrac{0-4^{2} _{1}}{-2g}= \dfrac{4 _{1}^{2}}{2g}= \dfrac{t _{1}^{2} g^{2} }{2g}= \dfrac{t _{1}^{2} g}{2}$
Similarly $h _{2}= \dfrac{t _{2}^{2} 9}{2} \Rightarrow \dfrac{h}{h _{2}}= \left( \dfrac{t _{1}}{t _{2}} \right)^{2} = \left( \dfrac{6}{9} \right)^{2}= \dfrac{4}{9}$
$\Rightarrow  (D)$

Mohit travelled $225$ km in $\displaystyle 4\frac {1}{2}$  hours. Raj travelled the same distance at an average speed that was $10$ km per hour faster than Mohit's average speed. What was the number of hours it took Raj to travel $225$ km ?

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. $\displaystyle 2\frac {1}{4}$

  2. $\displaystyle 3\frac {1}{4}$

  3. $\displaystyle 3\frac {3}{4}$

  4. $\displaystyle 4\frac {2}{5}$

Show answer
Correct Option: C
Explanation:

Using the formula, $Speed=\frac { Distance }{ time } $
Mohit speed=
${ S } _{ m }=\frac { 225 }{ 4.5 } =50km/h$
${ S } _{ r }=50+10=60km/h$
Distance=225km
$Time=\frac { Distance }{ Speed } =\frac { 225 }{ 60 } =3\frac { 3 }{ 4 } hr$
Answer (C) $3\frac { 3 }{ 4 } hr$

A grasshopper can jump up to $91.4cm$ in a single jump. If the grasshopper jumped $731.2cm$ altogether, how many jumps did it make?

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. $17$

  2. $8$

  3. $9$

  4. $12$

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Correct Option: B
Explanation:

$\Rightarrow$  Distance covered by grasshopper in one jump = $91.4\,cm$.

$\Rightarrow$  Total distance covered by grasshopper = $731.2\,cm$

$\Rightarrow$  Jumps required to covered total distance = $\dfrac{731.2}{91.4}$

$\therefore$  Jumps required to covered total distance = $8$ 

Two trains from two places start running in the opposite direction and reach the destination at the midpoint after  $3\dfrac{1}{3}$ hours and $4\dfrac{4}{5}$  hours. If the speed of the first train is $80$ km/hr, then the speed of the 2nd train (in km/hr) is

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. $64\dfrac{2}{3}$

  2. $66\dfrac{2}{3}$

  3. $55\dfrac{5}{9}$

  4. $75$

Show answer
Correct Option: C
Explanation:
$\text{Let speed of second train is x km/hr}$

$\text{if both train reaches the destination at the mid point then,}$

$\text{distance left to be covered by A = distance covered by B }.......(i)$
and
$\text{Distance = speed  x time taken}$

from (i)
$x\times\dfrac{24}{5}= 80 \times \dfrac{10}{3}$
$x=55\dfrac{5}{9}$

option c is correct