Question

a stone dropped from a height keeps on increasing its speed. Every second its speed increases by 10 metres per second. How much distance will the stone fall between 1.5 and 2.5 second, after being dropped with zero speed? Use the idea of average speed

Answer 1

A stone dropped from a height keeps on increasing its speed. Every second its speed increases by 10 metres per second. The distance that the stone falls between 1.5 seconds and 2.5 seconds is 20 m

Explanation:

The rate of change of speed = 10 m/s/s

Therefore, acceleration

a = 10 m/s²

Using the second equation of motion

$s=ut+\frac{1}{2}at^2$

The distance travelled by the stone in 1.5 seconds

$s_{1.5}=0\times 1.5+\frac{1}{2}\times 10\times 1.5^2$

$\implies s_{1.5}=5\times 2.25$

$\implies s_{1.5}=11.25$ m

Similarly

The distance travelled by the stone in 2.5 seconds

$s_{2.5}=0\times 2.5+\frac{1}{2}\times 10\times 2.5^2$

$\implies s_{2.5}=5\times 6.25$

$\implies s_{2.5}=31.25$ m

Therefore, the distance travelled between 1.5 and 2.5 seconds

$d=s_{2.5}-s_{1.5}$

$\implies d=31.25-11.25$ m

$\implies d=20$ m

Alternatively,

Using the concept of average velocity

From the first equation of motion

$v=u+at$

Velocity at the end of 1.5 seconds

$v_{1.5}=0+10\times 1.5$

$\implies v_{1.5}=15$ m/s

Similarly. velocity at the end of 2.5 seconds

$v_{2.5}=25$ m/s

Average velocity in the interval 1.5 to 2.5 seconds (i.e. in 1 second)

$v=\frac{v_{1.5}+v_{2.5}}{2}$

$\implies v=\frac{15+25}{2}$

$\implies v=20$ m/s

Therefore, distance travelled in 1 second (2.5 seconds - 1.5 seconds = 1 second)

$\text{Distance}=\text{Speed}\times\text{Time}$

or, $d=20\times 1$ m

$\implies d=20$ m

Hope this answer is helpful.

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