Question

a cyclist moving on a straight road covers a. distance 6km with speed 4m/s and 4km with speed 2.5m/s the average speed of cyclist is​

Answer 1

$\large{ \red {\underline{ \green{ \clubs \rm \red{Given:-}}}}}$

$Distance = 6km \: and \: 4km \\ \:</p><p>Speed = 4 {km \: s}^{ - 1} and \: 2.5km \: {s}^{? - 1}$

$\large{ \red{ \underline{ \green{ \clubs{ \red{To \: Find:-}}}}}}$

$The \: Average \: speed \: of \: the \: cyclist$

$\large{ \red{ \underline{ \green{ \clubs{ \red{Solution:-}}}}}}$

$</p><p> \blue{Average Speed = \frac{Total Distance}{ Total Time\: } }$

${ \: Distance}^{1} = 6km = 6000m$

${ \:Distance }^{2} = 4km = 4000m$

$Total \: Distance = 6000m + 4000m = 10000m$

$\blue{Time = \frac{ Distance \: }{ Speed \: \: } }$

${ \:Time}^{1} = \cancel\frac{6000 \: m}{4 \: m \: {s}^{ - 1} }= 1500 \: s$

${ Time }^{2} = \cancel\frac{4000 \: m}{2.5 \: m \: {s}^{ - 1} } = 1600 \: s$

$Total \: Time \: = 1500 \: s + 1600 \: s \\ \: \: \: \: \: \: \: \: = 3100 \: s$

$\blue{Average speed = \frac{ Total Distance \: }{ Total Time } } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \cancel \frac{10000 \: m}{3100 \: s} = 3.22 \: m \: {s}^{ - 1}$

$\green {\therefore{Average \: Speed \: of \: cyclist}} \: \: \\ = \green{ 3.22 \: m \: {s}^{ - 1} }$

Answer 2

SOLUTION

GIVEN

A cyclist moving on a straight road covers a. distance 6km with speed 4m/s and 4km with speed 2.5m/s

TO DETERMINE

The average speed of cyclist

EVALUATION

Here it is given that a cyclist moving on a straight road covers a. distance 6km with speed 4m/s and 4km with speed 2.5m/s

For distance of 6 km

Speed = 4 m/s

Distance covered = 6 km = 6000 m

Time taken

$\displaystyle \sf{ = \frac{6000}{4} \: \: sec}$

$\displaystyle \sf{ =1500 \: \: sec}$

For distance of 4 km :

Speed = 2.5 m/s

Distance covered = 4 km = 4000 m

Time taken

$\displaystyle \sf{ = \frac{4000}{2.5} \: \: sec}$

$\displaystyle \sf{ = 1600 \: \: sec}$

Calculation of average speed :

Total distance covered

= 6 km + 4 km

= ( 6000 + 4000 ) m

= 10000 m

Total time taken

= ( 1500 + 1600 ) sec

= 3100 sec

Hence the required average speed

$\displaystyle \sf{ = \frac{Total \: distance \: covered }{Total \: time \: taken } \: \: }$

$\displaystyle \sf{ = \frac{10000 }{3100 } \: \: } \: \: m/s$

$\displaystyle \sf{ = 3.23 } \: \: m/s \: \: (approx)$

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