Question

plzzzz help me..... ​

Answer 1

${\mathfrak{\underline{\orange{\:\:\: Given:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{ x = (1 + 2t + 3t^{2} )m}$

$\:\:\:\:\bullet\:\:\:\sf{Time = 4 \: sec }$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{Velocity \: after \: 4 \: sec\:( v_{t \ = \ 4}) }$

${\mathfrak{\underline{\red{\:\:\: Calculation:-\:\:\:}}}}$

☯ Finding velocity of particle after 4s

$\dashrightarrow\:\: \sf{ x = 1 + 2t + 3 {t}^{2} }$

$\dashrightarrow\:\: \sf{ \dfrac{dx}{dt} = 1 + 2t + {3t}^{2} }$

$\dashrightarrow\:\: \sf{v = 2 + 3 \times 2 \times t }$

$\dashrightarrow\:\: \sf{v = 2 + 6t }$

$\dashrightarrow\:\: \sf{ v = 2 + 6 \times 4 }$

$\dashrightarrow\:\: \sf{ v = 2 + 24 }$

$\dashrightarrow\:\: {\boxed{\sf{ v = 26 \: m/s }}}$

$\star\:\underline{\boxed{\sf{ Velocity \: of \: particle \: after \: 4 \: sec \: is \: {\purple{26 \: m/s}}}}}$

Answer 2

Answer:

The position of a particle moving along x-axis is given by

x=(−2t

3

+3t

2

+5)m

Velocity, v=

dt

dx

=0

−6t

2

+6t=0. . . . . . .(1)

6t(−t+1)=0

−t+1=0

t=1sec

Acceleration, a=

dt

dv

a=−12t+6

a∣

t=1sec

=−12×1+6

a∣

t=1sec

=−6m/s

2

Step-by-step explanation:

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