Question

Ram is travelling down a road A and covers a distance of 300 m. After that he takes a right turn and covers a distance of 400 m on road B. Find out the total distance and displacement covered by him. If he takes 10 minutes for road A and 15 minutes for road B, Find out the speed for each road and the average speed for the whole journey.also draw the diagram. ​

Answer 1

Answer:

Explanation:

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Answer 2

Given :-

Ram is travelling down a road A and covers a distance of 300 meters .

After that he takes a right turn and covers a distance of 400 meters on road B .

Required to find :-

• Total distance

• Total displacement

• Speed on each road

• Average speed

Solution :-

Given data :-

Ram is travelling down a road A and covers a distance of 300 meters .

After that he takes a right turn and covers a distance of 400 meters on road B .

Time taken to cross road A is 10 minutes .

Time taken to cross road B is 15 minutes .

we need to find the total distance , total displacement , speed on each road , Average speed .

So,

From the given data we can conclude that ;

Distance covered on road A ( d1 ) = 300 meters

Distance covered on road B ( d2 ) = 400 meters

Time taken to cross road A ( t1 ) = 10 minutes

Time taken to cross road B ( t2 ) = 15 minutes

Now,

Let's find the total distance ;

Total distance = Distance covered on road A + Distance covered on road B

=> Total distance = d1 + d2

=> Total distance = 300 m + 400 m

=> Total distance = 700 meters

Now,

Let's find the total displacement .

As we know that the displacement is the shortest path between any two places .

So, here the diagonal is the shortest path .

We need to use Pythagorean theorem ;

Before using it let let's convert the d1 and d2 from meters to kilometers in order to simplify our calculations .

This implies ;

1 meter = 1/1000 kilo meters

300 meters = ? km

=> 300/1000

=> 0.3 km

400 meters = ? km

=> 400/1000

=> 0.4 km

Using the Pythagorean rule ;

$\tt({d}_{1} {)}^{2} + ({d}_{2} {)}^{2} = (displacement {)}^{2}$

( 0.3 )² + ( 0.4 )² = (displacement)²

0.09 + 0.16 = displacement²

0.25 = displacement²

displacement² = 0.25

displacement = √0.25

displacement = 0.5 km

Hence,

Total displacement = 0.5 km ( or 500 meters )

Now,

Let's find the speed on each road

Road A ;

Distance = 300 meters

Time = 10 minutes

Since, distance in meters let's convert time into seconds

1 minute = 60 seconds

10 minutes = ?

=> 60 x 10

=> 600 seconds

This implies ;

Speed on the road A = 300/600

Speed on the road A = 1/2

Speed on the road A = 0.5 m/s

Hence,

Speed on the road A = 0.5 m/s

Similarly,

Road B ;

Distance = 400 meters

Time = 15 minutes

=> 15 x 60

=> 900 minutes

This implies ;

Speed on the road B = 400/900

Speed on the road B = 4/9

Speed on the road B = 0.444--

Speed on the road B = 0.4 m/s ( Approximately )

Hence,

Speed on the road B = 0.4 m/s ( approximately )

Now,

Let's find the Average speed

Using the formula ;

$\boxed{ \tt{ \red{average \: speed = } \green{ \dfrac{total \: distance}{total \: displacement}}}}$

Total distance = d1 + d2

=> 0.3 + 0.4

=> 0.7 km

Total time taken = t1 + t2

=> 10 + 15

=> 25 minutes

Here, convert the total time taken from minutes to Hours

Since,

1 minute = 1/60 hours

25 minutes = 25/60 hours

This implies ;

Average speed = 0.7 / 25/60

Average speed = 0.7 x 60/25

Average speed = 42/25

Average speed = 1.68 km/hr

Hence,

Average speed = 1.68 km/hr

Diagram :-

$\setlength{\unitlength}{30} \begin{picture}(6, 6) \put(2, 2){\line(0, 1){4}}\put(2, 2){\line( - 1, 0){4}}\put(2, 6){\line( - 1, - 1){4}} \put(2, 6){\circle*{0.2}}\put(2, 2){\circle*{0.2}}\put( - 2,2 ){\circle*{0.2}}\put(2.2, 6){ \bf Initial \: point }\put( - 2.2, 1.5){ \bf Final \: point }\put(2.2, 4){ \bf Road \: A }\put( - 1.5,2.2 ){ \bf Road \: B }\put(2.2, 3.5){ \bf 300 \: meters }\put( - 0, 2.2){ \bf 400 \: meters }\put( - 2,5){ \bf displacement }\put( - 2, 4.5){ \bf 500 \: meters }\put(0, - 0){\boxed{ \bf @MisterIncredible }}\end{picture}$