Question

A body is moving along a straight line. Its distance $X_t$ from a point on its path at a time $t$ after passing that point is given by $X_t=8t^2-3t^3$, where $X_t$ is in meter and $t$ is in second. The correct statement(s) is/are:

• A. Average speed during the interval $t=0$ s to $t=4s$ is $20.21{ms}^{-1}$
• B. Average velocity during the interval $t=0$ s to $t=4s$ is $-16{ms}^{-1}$
• C. The body starts from rest and at $t=\frac{16}{9}s$ it reverses its direction of motion at $x_t=8.43m$ from the start
• D. It has an acceleration of $-56{ms}^{-2}$ at $t=4s$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Average speed during the interval $t=0$ s to $t=4s$ is $20.21{ms}^{-1}$
B The body starts from rest and at $t=\frac{16}{9}s$ it reverses its direction of motion at $x_t=8.43m$ from the start
C Average velocity during the interval $t=0$ s to $t=4s$ is $-16{ms}^{-1}$
D It has an acceleration of $-56{ms}^{-2}$ at $t=4s$

$\Rightarrow X_t=8t^2-3t^3\Rightarrow \frac{d{X}_t}{dt}=V=16t-9t^2\Rightarrow \frac{d^2{X}_t}{{dt}^2}=a=16-18t$
Now,
$X_t=8t^2-3t^{3}Att=4seconds,X_t=-64metersAveragevelocity=\frac{-64}{4}=-16$
Now,
$\Rightarrow \frac{d{X}_t}{dt}=V=16t-9t^2\Rightarrow V=t(16-9t)\Rightarrow V=0att=0andt=\frac{16}{9}seconds.\Rightarrow Theobjectchangesitsvelocityinoppositedirectionatt=\frac{16}{9}seconds.$
Now,
$\Rightarrow \frac{d^2{X}_t}{{dt}^2}=a=16-18tAtt=4seconds,\Rightarrow a=16-(18\times 4)=-56$
Now,
$TotalDistancetravelledbyobject:betweent=0tot=\frac{16}{9}seconds\Rightarrow d=8.42\therefore Totaldistancetravelledbetweent=0tot=4secondsis\Rightarrow 64+(2\times 8.42)=80.84meters.\Rightarrow Averagespeed=\frac{80.84}{4}=20.21$