Question

A particle moves in a straight line with acceleration 'a'at time 't' where a=-1/t².at time t=15 the particle has a velocity =3ms¹ then velocity at t=4​

Answer 1

Answer:

$\boxed{\mathfrak{Velocity \ at \ 4 \ s = 2.8 \ m/s}}$

Explanation:

Accelration w.r.t. time:

$\rm a = \dfrac{1}{ {t}^{2} }$

Rate of change of velocity is known as accelration:

$\rm \implies a = \dfrac{dv}{dt} \\ \\ \rm \implies \dfrac{dv}{dt} = \dfrac{1}{ {t}^{2} } \\ \\ \rm \implies \int\limits^v_3dv = \int\limits^{t = 4}_{t = 15} \dfrac{1}{ {t}^{2} } .dt \\ \\ \rm \implies v\Big|_ 3^v = - \frac{1}{t} \Big|_{15}^{4} \\ \\ \rm \implies v - 3 = ( - \dfrac{1}{4} - ( - \dfrac{1}{15}) ) \\ \\ \rm \implies v - 3 = \dfrac{1}{15} - \dfrac{1}{4} \\ \\ \rm \implies v - 3 = \dfrac{4 - 15}{60} \\ \\ \rm \implies v - 3 = - \dfrac{11}{60} \\ \\ \rm \implies v = 3 - \dfrac{11}{60} \\ \\ \rm \implies v = \dfrac{180 - 11}{60} \\ \\ \rm \implies v = \dfrac{169}{60} \: m {s}^{ - 1} \\ \\ \rm \implies v = 2.8 \: m {s}^{ - 1}$