Question

a place is moving with uniform acceleration. Its displacement at any instant t is given y
s=10t + 49t^
what us
inital velocity
velocity at t=3
the uniform acceleration ​

Answer 1

$\huge\sf{\underline{Solution}}$

★Displacement = (s)

$10t + 4.9 {t}^{2}$

★Velocity

$v = \frac{ds}{dt} \\ = > 10 + 2 \times 4.9t$

$\boxed{\sf{Inital\:velocity}}$

(velocity t=0)

$= > 10 + 0 \\ = > 10m.{s}^{ - 1}$

$\boxed{\sf{Velocity\:at\:t=3\:second}}$

$v = 10 + 2 \times4.9 \times 3 \\ = > 39.4m.{s}^{ - 1}$

$\boxed{\sf{Acceleration}}$

$a = \frac{dv}{dt} \\ = > \frac{d(10 + 9.8t)}{dt} \\ = > 0 + 9.8 \\ = > 9.8m. {s}^{ - 1}$

Answer 2

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$\huge\color{red}\star\star{Hey mate:}\star\star$

$\huge\color{blue}\star\star{hope it help..}\star\star$

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