Question

Motion of body along straight line is described by equation x= t^3 +4t^2 –2t +5. Find velocity and acceleration of body at t=4s. Find average velocity and average acceleration during time interval from t=0s to t=4s

Answer 1

Answer:-

$v = 78 m/s \\a = 32 m/s^2\\ A_{avg}= 20 m/s^2 \\ A_v = 30m/s$

Given :-

$\mathsf{x = t^3 + 4t^2 -2t + 5 }$

To find :-

The velocity and acceleration at t = 4s.

Average velocity and average acceleration during time interval t = 0s to t = 4s.

Solution:-

The relation between time and displacement is given by equation.

$x = t^3 + 4t^2 -2t +5$

• Now, Differentiate x with w. r. t to t .

→$\dfrac{dx}{dt}= \dfrac{d( t^3 + 4t^2 -2t +5)}{dt}$

→$\dfrac{dx}{dt} = \dfrac{d(t^3)}{dt}+ \dfrac{d(4t^2)}{dt}-\dfrac{d(2t)}{dt}+\dfrac{d(5)}{dt}$

→$v = 3t^2 + 4 \times 2t - 2 \times 1+0$

→$v = 3t^2 + 8t -2$

• at t = 4s

Velocity will be :-

→$v = 3(4)^2 + 8\times 4 -2$

→$v = 3 \times 16 + 32 -2$

→$v = 48 + 32 -2$

→$v = 78 m/s$

• at t= 0 s

initial velocity will be :-

→$u = 3 (0)^2 + 8 \times 0 -2$

→$u = -2 m/s$

Now, Average acceleration of the body is :-

→$A_{avg}= \dfrac{v-u}{t_f - t_i}$

→$A_{avg} = \dfrac{78-(-2)}{4 - 0}$

→$A _{avg} = \dfrac{80}{4}$

→$A_{avg} = 20 m/s^2$

• Differentiate v with respect to t.

→$\dfrac{dv}{dt} = \dfrac{d(3t^2 +8t-2)}{dt}$

→$\dfrac{dv}{dt}= \dfrac{d(3t^2)}{dt} + \dfrac{d(8t)}{dt} - \dfrac{d(2)}{dt}$

→$a = 3 \times 2 t + 8 \times 1 + 0$

→$a = 6t +8$

• at t = 4s.

→$a = 6 \times 4 + 8$

→$a = 24 +8$

→$a = 32 m/s^2$

hence,

Acceleration will be 32 m/s^2.

• at t = 0 s

The displacement of the particle will be :-

→$x = t^3 + 4t^2 -2t + 5$

→$x = (0) ^3 +4(0) ^2 -2\times 0 + 5$

→$x = 5 m$

• at t = 4s

→$x = (4) ^3 + 4(4) ^2 -2(4) +5$

→$x = 64 + 64 -8 +5$

→$x = 133-8$

→$x = 125m$

$x = 125 - 5$

$x = 120m$

Average velocity of the body will be :-

$A_v = \dfrac{\text{Total displacement}}{\text{Total time taken}}$

→$A_v = \dfrac{125-5}{0+4}$

→$A_v = \dfrac{120}{4}$

→$A_v = 30m/s$

hence,

Average velocity will be 30 m/s.

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

${\green{\therefore{\text{Velocity=78m/s}}}}$

${\green{\therefore{\text{Acceleration=32m/}}s^{2}}}$

${\green{\therefore{\text{Avg.\:Velocity=30m/s}}}}$

${\green{\therefore{\text{Avg.\:Acceleration=}}20m/s^{2}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{Given}} \\\\ {:\implies{x = {t}^{3} + 4 {t}^{2} - 2t + 5(position \: of \: a \: particle }} \\ \\ \red{\underline{To \: Find}} \\\\ { :\implies{Acceleration =? }} \\\\ { :\implies{Velocity =? }} \\\\ { :\implies{Average \: acceleration = ?}} \\\\ { :\implies{Average \:velocity = ?}}$

• According to the given question:

• for finding velocity if position of a particle is given so we differentiate position of a particle.

$:\implies x = {t}^{3} + 4 {t}^{2} -2t+5\\\\ :\implies \frac{dv}{dt} = 3 {t}^{2} + 8t - 2 \\ \\ :\implies at \: (t = 4s) = 3 \times {4}^{2} + 8 \times 4 - 2 \\\\ :\implies velocity = 3 \times 16 + 32 - 2 \\\\ { \green{\therefore velocity = 78m/s}}$

• for finding acceleration if we get velocity of a particle so again differentiate the velocity of particle.

$v=3{t}^{2}+8t-2\\\\ :\implies acceleration = \frac{da}{dt} = 6t+8 \\\\ \implies {acc}^{n} = 6t + 8 \\ \\ :\implies at \: (t = 4s) = 6 \times 4 + 8 \\\\ { \green{\therefore {acc}^{n} = 32 m/s^{2} }}$

• Avg velocity during t=0 to t=4s.

$:\implies Initial \: displacement(at \: t = 0) = {0}^{3} + 4 \times {0}^{2} - 2 \times 0 + 5 \\\\ :\implies u = 5m \\ \\ :\implies Final \: displacement (at \: t = 4) = {4}^{3} + 4 \times {4}^{2} - 2 \times 4 + 5 \\\\ \implies v = 125m \\ \\ \implies Avg \: velocity = \frac{v - u}{t2 - t1} \\\\ \implies Avg \: velocity = \frac{125 - 5}{4 - 0} \\\\ { \green{\therefore Avg \: velocity = 30 m/s }}$

• Avg acceleration during t=0 to t=4s

$:\implies Avg \: acceleration = \frac{ v_{f} - v_{i} }{ t_{f} - t_{i} } \\\\ \implies Avg \: acceleration = \frac{78 - ( - 2)}{4 - 0} \\\\ { \green{\therefore Avg \: acceleration = 20 m/{s}^{2} }}$