Question

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

Answer 1

Answer:

4.0 m/s

t

1

=

3

2

S

=

6

S

2

S

=4.5×t

2

+7.5t

3

⇒12t

2

=

2

S

We have t

2

=t

3

Total time t=t

1

+t

2

+t

3

=

6

S

+

24

S

+

24

S

=

4

S

Average of

t

S

=

4

S

S

=4m/s

Answer 2

The average velocity of the cyclist is 5.058m$s^{-1}$

Step-by-step explanation:

Distance travelled initially, s1= 6km

Velocity 1; V1= 4m$s^{-1}$

Distance travelled later, s2= 4 km

Velocity 2; V2= 2.5m$s^{-1}$

Solution:

The time taken to travel 6km  = V1÷ S1

=4÷6= 0.66 seconds

The time taken to travel 4km= V2÷ S2

= 2.5÷4= 0.625 seconds

Now, The formula for average speed calculation is = initial speed+ final speed÷ total time taken

Putting the values, we get:

The average speed= 4+2.5÷ 0.66+0.625

=6.5÷1.285

The average speed= 5.058m$s^{-1}$

Result:

Thus, the average velocity of the cyclist is 5.058m$s^{-1}$

(#SPJ3)