Question

An athlete starts from rest with an acceleration 2 ms² for sometime and then decelerates with 3 ms² and finally comes to rest .If the length of track is 100 m, then maximum speed attained by athletr is

Answer 1

Let maximum velocity of athlete is v.
initial velocity of athlete , u = 0
distance travelled by athlete with an acceleration 2 m/s² is s
use formula, v² = u² + 2as
v² = 0 + 2 × 2 × s = 4s .......(1)

athlete decelerates and comes to rest.
so, final velocity of athlete, v' = 0
initial velocity of athlete , u = v
distance travelled by athlete is (100 - s)
use formula, v² = u² + 2as
0 = v² - 2 × 3 × (100 - s)
v² = 6(100 - s) ..........(2)

from equations (1) and (2),
4s = 6(100 - s)
4s = 600 - 6s
4s + 6s = 600
10s = 600
s = 60 m

so, v² = 4s = 4 × 60 = 240
v = $\sqrt{240}$

Answer 2

HEY DEAR ... ✌️

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Let maximum velocity of athlete be V

Then , initial velocity of athlete , u = 0

Given , Distance travelled by an athlete with acceleration 2 m/s² is S

Acc. to the question ,

we know , v² = u² + 2as

v² = 0 + 2 × 2 × s = 4s ---------(1)

Athlete decelerates and comes to rest.

so, final velocity of athlete will be v' = 0

Initial velocity of athlete will be u = v

So , Distance travelled by an athlete is (100 - s)

Now , using formula , v² = u² + 2as

By putting values ,

0 = v² - 2 × 3 × (100 - s)

v² = 6(100 - s) -----------(2)

From above equations (1) and (2),

4s = 6(100 - s)

4s = 600 - 6s

4s + 6s = 600

10s = 600

S = 60 m

so, V^2 = 4s

4 × 60

240

√240


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HOPE , IT HELPS ... ✌️