Question

Please tell me the answer fast

Answer 1

Given :

• velocity of the body

V = t ³ - 6 t² + 10 t + 4

To find :

• acceleration of the body at

1. t = 0 sec
2. t = 1 sec
3. t = 5 sec

Solution :

we know that ,

$\displaystyle \bigstar \: \boxed{ \green{\mathtt{a = \frac{dv}{dt} }}}$

here ,

• a = acceleration
• v = velocity

Substituting the value of v in the above equation ,

$\implies\displaystyle\mathtt{a = \frac{d( {t}^{3} - 6 {t}^{2} + 10t + 4)}{dt} }$

$\implies\displaystyle\mathtt{a = 3{t}^{2} - 12 t + 10 }$

At, t = 0 sec

$\implies\displaystyle\mathtt{a_1 = 3({0}^{2} )- 12 (0) + 10 }$

$\boxed{ \red{ \mathtt{a_1 = 10 \frac{m}{ {s}^{2} } }}}$

At t = 1 sec

$\implies\displaystyle\mathtt{a_2 = 3({1}^{2}) - 12 \times 1 + 10 }$

$\implies\displaystyle\mathtt{a_2 = 3 - 12 + 10}$

$\displaystyle \boxed{ \red{\mathtt{a_2 = 1 \frac{m}{ {s}^{2} } }}}$

At t = 5 sec

$\implies\displaystyle\mathtt{a_3 = 3({5}^{2}) - 12 \times 5 + 10 }$

$\implies\displaystyle\mathtt{a_3 = 75 - 60 + 10}$

$\implies\displaystyle\mathtt{a_3 = 85 - 60}$

$\displaystyle \boxed{ \red{\mathtt{a_3 = 15 \frac{m}{ {s}^{2} } }}}$

Increasing order of acceleration

a 2 < a 1 < a 3

Hence the answer is a