Question

2. An athlete starting from rest picks up a velocity of 10 m/s over a distance of 100 m. Calculate (a) the we
acceleration generated by the athlete, and (b) the time in which he completes the race.





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Answer 1

☞ To Find :

• The acceleration of the athlete.

• The time taken to finish the the race.

☞ Given :

• Final velocity $\it{\rightarrow v = 10ms{-1}}$

• Intial velocity $\it{\rightarrow u = 0ms{-1}}$

• Distance covered $\it{\rightarrow s = 100m}$

☞ We Know :

• Second law of motion :

$\underline{\boxed{s = ut \pm \dfrac{1}{2}at^{2}}}$

↝ Where,

➝ s = Distance Covered

➝ u = Intial velocity

➝ v = Final velocity

➝ t = Time taken

➝ a = Acceleration produced

• Third law of motion :

$\underline{\boxed{v^{2} = u^{2} \pm 2as}}$

↝ Where,

➝ s = Distance Covered

➝ u = Intial velocity

➝ v = Final velocity

➝ a = Acceleration produced

☞ Concept :

The Initial velocity is taken as 0 , as the athlete is starting from rest ...

Hence , we can form the special Equation ,when initial velocity is 0.

• For the second law of motion.

$\mathtt{s = ut \pm \dfrac{1}{2}at^{2}}$

Putting the initial velocity as 0 ,we get :

$\mathtt{\Rightarrow s = 0 \times t \pm \dfrac{1}{2}at^{2}}$

$\mathtt{\Rightarrow s = \pm \dfrac{1}{2}at^{2}}$

So the Equation formed for motion , when Intial velocity is 0 is :-

$\mathtt{\therefore s = \pm \dfrac{1}{2}at^{2}}$

• For third law of motion :

$\mathtt{v^{2} = u^{2} \pm 2as}$

Putting the initial velocity as 0 ,we get :

$\mathtt{\Rightarrow v^{2} = 0^{2} \pm 2as}$

$\mathtt{\Rightarrow v^{2} = \pm 2as}$

$\mathtt{\Rightarrow v^{2} = \pm 2as}$

So the Equation formed for motion , when Intial velocity is 0 is :-

$\mathtt{\therefore v^{2} = \pm 2as}$

The acceleration will be positive , as the athlete is accelerating from rest to motion .

☞ Solution :

→ For Finding the Acceleration :

Using the third law motion : (Where ,u = 0)

$\mathtt{v^{2} =2as}$

(Here, a Acceleration is Taken as postive as the athlete is accelerating from rest to motion.)

Putting the value in the equation ,we get :

$\mathtt{\Rightarrow 10^{2} = 2a \times 100}$

$\mathtt{\Rightarrow 100 = 200a}$

$\mathtt{\Rightarrow \dfrac{100}{200} = a}$

$\mathtt{\Rightarrow \dfrac{\cancel{100}}{\cancel{200}} = a}$

$\mathtt{\Rightarrow 0.5 ms^{-2} = a}$

Hence ,the acceleration produced by the athlete is $\mathtt{0.5 ms^{-2} = a}$

→ For finding the time taken :

Using the second law of motion : (Where ,u = 0)

$\mathtt{s = \dfrac{1}{2}at^{2}}$

(Here, a Acceleration is Taken as postive as the athlete is accelerating from rest to motion.)

Putting the value in the equation ,we get :

$\mathtt{\Rightarrow 100 = \dfrac{1}{2} \times 0.5 \times t^{2}}$

$\mathtt{\Rightarrow 2 \times 100 = 0.5 \times t^{2}}$

$\mathtt{\Rightarrow 200 = 0.5 \times t^{2}}$

$\mathtt{\Rightarrow \dfrac{\cancel{200}}{\cancel{0.5}} = t^{2}}$

$\mathtt{\Rightarrow 400 = t^{2}}$

$\mathtt{\Rightarrow \sqrt{400} = t}$

$\mathtt{\Rightarrow 20 s = t}$

Hence , the time taken to complete the race is 20 s.

Extra Information :

• First law of motion :

$\mathtt{v = u + at}$

• Formula for nth second :

$\mathtt{s_{n} = u + \dfrac{1}{2}a(2n - 1)}$

↝ Where,

➝ s = Distance Covered

➝ u = Intial velocity

➝ v = Final velocity

➝ t = Time taken

➝ a = Acceleration produced