Question

The velocity of a moving particle is given by
the equation v = (5-t2)ms. The average
acceleration of the particle between the 2nd
and 3 seconds is: -
1) -5ms
2) 2ms
3) 4ms²
4) -4ms2​

Answer 1

I don't know the answer, sorry, sorry

Answer 2

Answer:

The correct option is 1) - 5 $ms^{-2}$

Explanation:

Here we have been given that the velocity of the moving particle is demonstrated by the equation v = 5 - $t^{2}$ m/s

Now we have to find the average acceleration of the particle between the time of 2nd and 3rd second.

Now as we know that the average acceleration of the given moving particle is the ratio of the change in the velocity of the particular particle and the change in time.

Therefore we will first calculate the velocity $v_{1}$ of the moving particle at    $t_{1}$ = 2 seconds as below:

v =  $5 -t^{2}m/s$

⇒ $v_{1} =(5 - 2^{2} )m/s$

⇒ $v_{1} = (5 - 4 )m/s$

⇒ $v_{1} = 1m/s$

Similarly, at $t_{2}$ = 3 seconds the velocity $v_{2}$ of the particle is as below:

⇒ $v_{2} = (5 - 3^{2}) m/s$

⇒ $v_{2} = (5 - 9) m/s$

⇒ $v_{2} = -4m/s$

Now the average acceleration of the particle is as below:

Average acceleration = $\frac{v_{2}-v_{1}}{t_{2}-t_{1}}$

= $\frac{(-4-1 )m/s}{(3-2) s}$

= -5 $ms^{-2}$

Therefore the average acceleration of the given moving particle is

- 5 $ms^{-2}$

Hence the correct option is option 1) - 5 $ms^{-2}$