Question

If v=(t^(2)-4t+10^(5))m/s where t is in second. Find acceleration at t=1 sec.

Answer 1

Given

Velocity of the particle is defined as :

$\sf \: v(t) = { {t}^{2} - 4t + {10}^{5} } \: \: \: {ms}^{ - 1}$

Differentiating velocity w.r.t time gives acceleration of the particle :

$\sf \: a(t) = \dfrac{dv}{dt} \\ \\ \longrightarrow \: \sf \: a(t ) = \dfrac{d( {t}^{2} - 4t + {10}^{5}) }{dt} \\ \\ \longrightarrow \ \boxed{ \boxed{ \sf \: a = (2t - 4) \: \: {ms}^{ - 2} }}$

Putting t = 1s,

$\longrightarrow \: \sf \: a =2 \times 1 - 4 \\ \\ \longrightarrow \: \underline{ \boxed{ \sf \: a = - {2 \: \: ms}^{ - 2} }}$

Acceleration of the particle at t = 1s is - 2 m/s²

Answer 2

$\huge\bold\green{Answer:-}$

Velocity of a particle =

$vt = t { }^{2} - 4t + {10}^{5}{ms }^{ - 1}$

Acceleration of the particle =

$a(t) = dv \div dt$

$a = (2t - 4) {ms}^{2}$

Putting t = 1s we get,

$a = 2 \times 1 - 4$

$a = - 2 {ms}^{2}$

Therefore acceleration of the particle is -2ms^2