Question

A scooter travelling at 10 m/s speed up to 20 m/s in 4 sec. Find the acceleration of scooter​

Answer 1

Answer:

Acceleration of scooter is 2.5 m/s ^2

Answer 2

→ To Find :

The Acceleration of the scooter ....

→ Given :

• Initial velocity $\rightarrow u = 10ms^{-1}$

• Final velocity $\rightarrow v = 20ms^{-1}$

• Time taken $\rightarrow t = 4 s$

→ We Know :

First law of motion :

${\boxed{\mathtt{v = u + at}}}$

Where,

• v = Final velocity
• u = Initial velocity
• t = Time taken
• a = Acceleration

→ Concept :

• The acceleration will be taken as postive , as the scooter is accelerating from Intial velocity to final velocity...

• The acceleration will be constant in this case..

→ Solution :

• v = $20 ms^{-1}$
• u = $10 ms^{-1}$
• t = $4 s$

Formula :

$\mathtt{v = u + at}$

Putting the value in the formula ,we get :

$\mathtt{\Rightarrow 20 = 10 + a \times 4}$

$\mathtt{\Rightarrow 20 = 10 + 4a}$

$\mathtt{\Rightarrow 20 - 10 = 4a}$

$\mathtt{\Rightarrow 10 = 4a}$

$\mathtt{\Rightarrow \dfrac{10}{4} = a}$

$\mathtt{\Rightarrow \dfrac{\cancel{10}}{\cancel{4}} = a}$

$\mathtt{\Rightarrow 2.5 ms^{-2} = a}$

Hence ,the acceleration of the scooter is $\mathtt{2.5ms^{-2}}$

→ Additional information :

Second law of motion :

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

Third law of motion :

$v^{2} = u^{2} + 2as$

Formula for nth second :

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

Where,

• v = Final velocity

• u = Initial velocity

• t = Time taken

• a = Acceleration

• s = displacement

• n = nth second