Question

a particle is moving in a straight line.Its displacement at any instant 't' is given by x= 10t + 15t³,where x is in meter and t is in second.Find (1) the average acceleration in the time interval t=0 to t= 2 second and (2) instantaneous acceleration at t= 2 second?

Answer 1

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

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

X = 10t + 15t³.

t = 2 seconds.

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

$\large{\bold{X = 10t + 15t^3}}$

First Finding velocity.

And For that we need to differentiate.

$\large{\bold{v = \frac{dx}{dt}}}$

Substituting the values.

$\large{v = \frac{d(10t + 15t^3)}{dt}}$

That will give

$\large{\boxed{ v = 10 + 45t^2}}$

For average acceleration :-

$\large{\bold{a = \frac{v_2 - v_1}{t_2 - t_1}}}$

Now,

$\large{ \implies a = \frac{[10 + 45 \times 4] - [10 + 0] }{2 - 0}}$

Substituting t = 2 seconds in the ${v_2}$ and t = 0 seconds in ${v_1}$

Solving

$\large{ \implies a = \frac{\cancel{10} + 180 - \cancel{10}}{2}}$

$\large{ \implies a = \frac{180}{2}}$

$\huge{\boxed{\boxed{a = 90 \: m/s^2 }}}$

So,the average acceleration is 90 m/s²

For instantaneous acceleration :-

Differentiating velocity.

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

Substituting the values.

$\large{ \implies a = \frac{d(10 + 45t^2)}{dt}}$

$\large{ \implies a = 90t}$

As t = 2 seconds.

a = 90 × 2

$\huge{\boxed{\boxed{a = 180 \: m/s^2}}}$

So,the instantaneous acceleration is 180 m/s².

Answer 2

$\huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

From the Question,

The position "x" of the particle is defined by the relation,

$\large{\sf{x = 10t + 15t {}^{3} }}$

Differentiating x w.r.t to to,we get velocity of the particle:

$\sf{v = \frac{dx}{dt} } \\ \\ \implies \: \sf{v = \frac{d(10t + 15t {}^{3} )}{dt} } \\ \\ \implies \: \green{\huge{\boxed{\boxed{\sf{v = 10 + 45t {}^{2} }}}}}$

Differentiating v w.r.t to t,we get the acceleration of the particle:

$\sf{a = \frac{dv}{dt} } \\ \\ \implies \: \sf{a = \frac{d(10 + 45t {}^{2}) }{dt} } \\ \\ \implies \: \green{ \huge{\boxed{ \boxed{ \sf{a = 90t}}}}}$

Putting t = 0,we get:

$\sf{{a}_{I} = 90(0)} \\ \\ \rightarrow \ \large{\sf{{a}_{I} = 0m{s}^{-2}}}$

Putting t = 2,we get:

$\sf{{a}_{f} = 90(2)} \\ \\ \rightarrow \ \large{\sf{{a}_{f} = 180m{s}^{-2}}}$

Now,

Average Acceleration

$\sf{ \bar{a} \: = \frac{180 + 0}{2} } \\ \\ \implies \: \huge{\sf{ \bar{a} = 90ms {}^{ - 2} }}$

The average acceleration of the particle during the time interval is 90m/s²

• The acceleration of the particle at t = 2s is 180m/s²