Question

A particle moves along a straight line and its velocity depends on time $t$as $v = 4t - {t^2}$. Here $v$is in meters per second and $t$is in seconds. Then for the first 5 seconds:

$A.$ Magnitude velocity of average velocity is $\dfrac{5}{3}m$per $s$
$B.$Average speed is $\dfrac{{13}}{5}$ $m$per $s$
$C.$ Average speed is $\dfrac{{11}}{5}m$per s
$D.$ Average acceleration is $- 1m$per ${s^2}$

Answer 1

Hint: Here we will proceed by calculating the magnitude velocity, then average speed after that by calculating final velocity and initial velocity we can calculate average acceleration.

Step By Step Answer:
Formula used:

Average acceleration is${A_{avg}} = \dfrac{{vf - vu}}{{\vartriangle t}}$

Average Speed (S)= $\dfrac{{dis\tan cetravelled\left( s \right)}}{{timetaken\left( t \right)}}$
Average velocity that is $vavg = \dfrac{s}{{\vartriangle t}}$

Here it is given that,

$v = 4t - {t^2}$

Where $'v'$is velocity of body in meter per second
And, $'t'$is time in seconds.

In order to calculate distance ‘s’, we need to Integrate the value of velocity $'v'$ with the function of time, $'t'$over the limit 0 to 5 sec.

So,

$S = \int\limits_0^5 {vdt}$
$\Rightarrow S = \int\limits_0^5 {4t - {t^2}dt}$ ….. (1)
$\Rightarrow S = \int\limits_0^5 {4td - \int\limits_0^5 {{t^2}} } dt$
$\Rightarrow S = 4\int\limits_0^5 {tdt} - \left[ {\dfrac{{{t^3}}}{3}} \right]_0^5$
$\Rightarrow S = 4\left[ {\dfrac{{{t^2}}}{2}} \right]_0^5 - \left[ {\dfrac{{{5^3}}}{3} - \dfrac{{{0^3}}}{3}} \right]$

$\Rightarrow S = 2\left[ {{5^2} - {0^2}} \right] - \left[ {\dfrac{{125}}{3} - 0} \right] \\\ \Rightarrow S = 2\left( {25} \right) - \dfrac{{125}}{3} \\\ \Rightarrow S = 50 - \dfrac{{125}}{3} \\\ \Rightarrow S = \dfrac{{50 - 125}}{3} \\\ = \dfrac{{25}}{3}m \\\$

Therefore, Distance ‘S’ $= \dfrac{{25}}{3}m$
And, as we know that

Average velocity $Vavg = \dfrac{{Dis\tan ce\left( s \right)}}{{Time\left( t \right)}}$

$= \dfrac{{\dfrac{{25}}{3}}}{5} \\\ = \dfrac{{25}}{3} \times \dfrac{1}{5} \\\$

$= \dfrac{5}{3}$ meter per second.

Again, average speed is given by,

$= \dfrac{{Dis\tan ce\operatorname{cov} ered}}{{Timetaken}} \\\ = \dfrac{{Dis\tan ce}}{{\vartriangle t}} \\\$

Now, Distance travelled in case of Average speed is calculated by integrating over the function of time from limits 0 to 4 with the sum of integrating speed over the function of time from limit 4 to 5.

Distance, $S = \int\limits_0^4 {vdt} + \int\limits_4^5 { - vdt}$

$= \int\limits_0^4 {4t - {t^2}dt - \int\limits_4^5 {4t - {t^2}} } dt$ ….. (By using equation 1)

= \left[ {\dfrac{{4{t^2}}}{2} - \dfrac{{{t^3}}}{3}} \right]_0^4 - \left[ {\dfrac{{4{t^2}}}{2} - \dfrac{{{t^3}}}{3}} \right]_4^5 \\\ = \left[ {\dfrac{{4{{\left( 4 \right)}^2}}}{2} - {{\dfrac{{\left( 4 \right)}}{3}}^3}} \right] - \left[ {\dfrac{{4{{\left( 0 \right)}^2}}}{2} - \dfrac{{{{\left( 0 \right)}^2}}}{3}} \right] - \left\\{ {\left[ {\dfrac{{4{{\left( 5 \right)}^2}}}{2} - \dfrac{{{{\left( 5 \right)}^3}}}{3}} \right] - \left[ {\dfrac{{4{{\left( 4 \right)}^2}}}{2} - \dfrac{{{{\left( 4 \right)}^3}}}{3}} \right]} \right\\} \\\

= \left[ {\dfrac{{64}}{2} - \dfrac{{64}}{3}} \right] - \left[ 0 \right] - \left\\{ {\left( {50 - \dfrac{{125}}{3}} \right) - \left( {32 - \dfrac{{64}}{3}} \right)} \right\\}
$= \left[ {\dfrac{{32}}{7} - \dfrac{{64}}{3}} \right] - \left[ {50 - \dfrac{{125}}{3} - 32 + \dfrac{{64}}{3}} \right]$

$= 32 - \dfrac{{64}}{3} - 50 + \dfrac{{125}}{3} + 32 - \dfrac{{64}}{3} \\\ = 14 - 1 \\\ = 13m \\\$

So, Average Speed (S)= $\dfrac{{dis\tan cetravelled\left( s \right)}}{{timetaken\left( t \right)}}$

$= \dfrac{{13}}{5}$ meter per second.
Now, again average acceleration is given by

${A_{avg}} = \dfrac{{vf - vu}}{{\vartriangle t}}$

$\Rightarrow {A_{avg}} =$ where ${a_{avg}}$is average acceleration
$vf =$ final velocity
$vu$= initial velocity
$\vartriangle t =$ time
$vf = 4t - {t^2}$ v

In case of velocity t will be 5 sec

$\therefore vf = 4\left( 5 \right) - {\left( 5 \right)^2} \\\ = 20 - 25 \\\$

$= - 5$ meter per second.

In case of initial velocity, t will be 0 sec.

$\therefore vi = 4\left( 0 \right) - \left( 0 \right) \\\ = 0 - 0 \\\ = 0 \\\$

Now,

$Aavg = \dfrac{{vf - vi}}{t} \\\ = \dfrac{{ - 5 - 0}}{5} \\\ = \dfrac{{ - 5}}{5} \\\$

     $ = - 1$ meter per second$^2$   

Thus, Average acceleration will be $- 1$ meter per second$^2$

Therefore, option A, B and D holds for this question.

Note: Whenever we come up with this type of question, where we are asked to find out the velocity of a particle. Then we first write down the formula for average velocity. After that we will calculate average speed. By calculating average speed, we can calculate average acceleration. Thus by solving questions step by step we will find our answer.