Question

If the velocity of a particle undergoing rectilinear motion
is given by v = (5 + 2t4) m/s, then the average
acceleration of particle between t = 2 s and t = 5 s is

Answer 1

Given:

The velocity of a particle undergoing rectilinear motion is given by v = (5 + 2t⁴) m/s.

To find:

Average acceleration between t = 2 sec and t = 5 sec.

Calculation:

Average acceleration is defined as the the change in velocity divided by the change in time.

$\therefore \: avg. \: a = \dfrac{\Delta v}{\Delta t}$

$= > \: avg. \: a = \dfrac{ v_{5} - v_{2} }{5 - 2}$

$= > \: avg. \: a = \dfrac{ \{5 + 2( {5}^{4} ) \} - \{5 + 2( {2}^{4}) \} }{5 - 2}$

$= > \: avg. \: a = \dfrac{ \{1255\} - \{37 \} }{3}$

$= > \: avg. \: a = 406 \: m {s}^{ - 2}$

So, final answer is:

$\boxed{ \sf{ \: avg. \: a = 406 \: m {s}^{ - 2} }}$

Answer 2

Question:-

If the velocity of a particle undergoing rectilinear motion is given by v = (5 + 2t4) m/s, then the average acceleration of particle between t = 2 s and t = 5 s is

Answer:-

Average acceleration is given by:-

$\dagger\sf{\boxed{\large{{A}_{avg}\:=\:\dfrac{{v}_{f}\:-\:{v}_{i}}{{t}_{f}\:-\:{t}_{i}}}}}$

Here, $\checkmark\sf{{v}_{f}\:=\:velocity\:at\:t\:=\:5}$

$\implies\sf{{v}_{f}\:=\:5\:+\:(2\:*\:{5}^{4}}$

$\implies\sf{{v}_{f}\:=\:1255}$

Here, $\checkmark\sf{{v}_{i}\:=\:velocity\:at\:t\:=\:2}$

$\implies\sf{{v}_{i}\:=\:5\:+\:(2\:*\:{2}^{4}}$

$\implies\sf{{v}_{f}\:=\:37}$

Here, $\checkmark\sf{{t}_{f}\:=\:5s}$

Here, $\checkmark\sf{{t}_{i}\:=\:2s}$

So putting the values in the formula,

$\sf{\large{{A}_{avg}\:=\:\dfrac{1255\:-\:37}{5\:-\:2}}}$

$\implies\sf{\large{{A}_{avg}\:=\:\dfrac{1218}{3}}}$

$\implies\sf{\large{{A}_{avg}\:=\:406m{s}^{-2}}}$

Ans. Average Acceleration = 406m/s²

Note:-

• Average Acceleration and Instantaneous acceleration are different. The formula for Instantaneous acceleration is $\sf{\overrightarrow{a}\:=\:\lim_{\Delta{t}\rightarrow0}{\dfrac{\Delta\overrightarrow{v}}{\Delta{t}}}\:=\:\frac{\text{d}\overrightarrow{v}}{\text{d}t}}$