Question

A car travels along the three sides of a equilateral triangle at the speeds of 200, 400 and 600 km/hr. Find the average speed of the car around the field? ( approximately)

Answer 1

Answer:

Let the side of the triangle be 'a'.

For first side:

Speed,u = 200 km/hr

So Time Taken, t1 = Distance/Speed=a/200 he

Similarly for second and third side,

t2=a/400 hr and t3=a/600 he

Now,

Avg. speed = Total distance covered/Total time taken

=3*a/t1+t2+t3

$= 3a \div (a \div 200 + a \div 400 + a \div 600)$

$\frac{3a}{ \frac{6a + 3a + 2a}{1200} }$

$= \frac{3a \times 1200}{11a}$

$= \frac{3600}{11}$

= 327.27 km/hr

Answer 2

Answer:

The average speed of the car around the field is 327.27 km/hr

Explanation:

Definition:
Equilateral triangle:

• An equilateral triangle is a triangle with the same length on all three sides.
• An equilateral triangle is also equiangular in Euclidean geometry; that is, all three internal angles are equivalent to each other and are each 60°.

Average speed:

• The overall distance traveled by the object in a given time frame is the average speed.
• A scalar quantity is an average speed.
• It is expressed by the magnitude and lacks direction.

Formula:

• Speed= Distance/ Time
• Average Speed= Total Distance/ Total Time

Solution:

Let "a" km be the length of each side of the equilateral triangle.

Thus, total distance covered= a+a+a= 3a km

Let the speed for all sides will be S1, S2, and S3 km/hr resp.

S1 = 200 km/hr

S2 = 400 km/hr

S3 = 600km/hr

Let the time take be T1, T2, and T3 seconds resp.

Substituting the values,

The avg speed changes to,

$=\frac{3a}{T1+T2+T3}$

$=\frac{3a}{\frac{a}{200} +\frac{a}{400} +\frac{a}{600} }$

$=\frac{3}{\frac{6+3+2}{1200}}$

$=\frac{3*1200}{11}$

$=\frac{3600}{11}$

= 327.27 km/hr

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