Question

If s=(2t+4t^2)m,find the values of 'v' and 'a' at t=0s,2s and 10s​

Answer 1

Answer:

(at t=0sec) v=2 m/s

(at t=2s) v=2+8t=2+8×2 =18 m/s

(at t=10s) v= 82 m/s

acceleration will be 8 m/s²

at all the given time.

Explanation:

s=2t+4t²

differentiating both sides

we get velocity

v=2+8t.

again differentiating both sides

we get acceleration

a=8.

bcoz acceleration is independent of time

therefore, acceleration will be constant.

now, we want velocity

so we will put the value of time in our velocity equation i.e; v=2+8t

so put t=0 , we get v=2 m/s

at t=2 sec , v= 2+8t = 2+8×2 = 2+16 = 18m/s

and at t=10sec , v=2+8×10 = 2+80 = 82m/s.

.. thanks.

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

$\large{\bold{S = 2t + 4t^2}}$

$\huge{\bold{\underline{Explanation:-}}}$

Finding the Velocity,

$\large{\bold{v = \dfrac{ds}{dt}}}$

Substituting the values,

$\large{v = \dfrac{d(2t + 4t^2)}{dt}}$

Differentiating,

$\large{\boxed{\boxed{v = 2 + 8t}}}$

Now, Differentiating velocity w.r.t time to get acceleration.

$\large{\bold{a = \dfrac{dv}{dt}}}$

Substituting the values,

$\large{a = \dfrac{d(2 + 8t)}{dt}}$

Differentiating,

$\large{\boxed{\boxed{a = 8 \: m/s^2}}}$

Now, Substituting time values,

Velocity:-

t = 0 Seconds,

$\large{v = 2 + 8t}$

$\large{v = 2 + 8 \times 0}$

$\large{\boxed{v = 2 \: m/s}}$

t = 2 Seconds,

$\large{v = 2 + 8t}$

$\large{v = 2 + 8 \times 2}$

$\large{\boxed{v = 18 \: m/s}}$

t = 10 Seconds,

$\large{v = 2 + 8t}$

$\large{v = 2 + 8 \times 10}$

$\large{\boxed{v = 82 \: m/s}}$

Acceleration:-

As acceleration is not dependent on time it will not change w.r.t time and will be constant throughout the journey.

Hence , the acceleration at t = 0, 2, 10 seconds will be 8 m/s².