Question

If the relation b/w acceleration and time for an object is given by A=2t+4t^2.
Calculate the position of object from the origin at t=4s.(At t=0, v=0, x=0)

Answer 1

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

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

A = 2t + 4t²

(Relation between acceleration and time)

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

A = 2t + 4t².

To find displacement it we should find out first displacement.

And for that we have to integrate it.

a = dv/dt

dv = a.dt

Integrating

${ \displaystyle \int dv = \displaystyle \int a.dt}$

Applying limits

${ \displaystyle \int_{0}^{v} dv = \displaystyle \int_{0}^{t} a.dt}$

on simplifying.

${v = \frac{2t^2}{2} + \frac{4t^3}{3}}$

${v = t^2 + \frac{4t^3}{3}}$

Again integrating to get displacement.

v = ds/dt

ds = v.dt

${ \displaystyle \int ds = \displaystyle \int v.dt}$

Applying limits

${ \displaystyle \int_{0}^{s} ds = \displaystyle \int_{0}^{t} v.dt}$

on simplifying.

${s = \frac{t^3}{3} + \frac{4t^4}{3 \times 4}}$

${s = \frac{t^3}{3} + \frac{t^4}{3}}$

Now, substituting t = 4 s.

${s = \frac{4^3}{3} + \frac{4^4}{3}}$

${s = \frac{64 + 256}{3}}$

${s = \frac{320}{3}}$

${S = 106.67 meters}$

$\huge{\boxed{\boxed{S = 106.67 meters}}}$

So, position of object will be 106.67 meters.

Answer 2

Answer:

106.6 m.

Explanation:

Given equation,

a = 2t + 4t²

Multiplying both the sides by dt.

a.dt = (2t + 4t²).dt

⇒ Now, Integrate both the sides of the Equation,

v = t² + 4/3t³

Now, again multiplying both the sides by dt.

v.dt = (t² + 4/3t³).dt

Again Integrating both the sides of the equation,

S = t³/3 + t⁴/3

At t = 4 seconds,

S = 64/3 + 256/3

⇒ S = 21.33 + 85.33

∴ S = 106.67 m.

Hence, Displacement of the body is 106.67 m at t = 4 seconds.

Hope it helps.