Question

an object moving in a straight line covers half of the distance with speed of 5m/s. The other half of the distance is covers in two equal time intervals with speed of 4m/s and 6m/s respectively. the average speed of the object is​

Answer 1

Answer:

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

Answer:

5m/s

Explanation:

Let S be the total distance travelled by particle.

Time taken by the particle to travel first half of the distance is

$t_{1} = \frac{S/2}{5} = \frac{S}{10}$

Let $t_{2}$ be the time taken by the particle to travel remaining half of the distance. Then,

$\frac{S}{2} = 4 \frac{t_{2} }{2} + 6\frac{t_{2} }{2}$

$= 5t_{2} \\= t_{2} = \frac{S}{10}$

The average speed of the particle is

$v_{av} = \frac{Total distance travelled}{Total time taken} \\ = \frac{S}{t_{1} + t_{2} } \\= \frac{S}{\frac{S}{10} + \frac{S}{10} } \\= \frac{1}{\frac{1}{10} + \frac{1}{10} } \\= \frac{1}{\frac{1+1}{10} } \\= \frac{1}{\frac{2}{10} } \\=\frac{10}{2} \\= 5m/s$