- Home
- Standard 11
- Physics
2.Motion in Straight Line
normal
A particle moves along a straight line in such a way that it’s acceleration is increasing at the rate of $2 m/s^3$. It’s initial acceleration and velocity were $0,$ the distance covered by it in $t = 3$ second is ........ $m$
A$27 $
B$9 $
C$3 $
D$1 $
Solution
The increase in acceleration is $2 m / s^{3}$
Time $t=3 s$
Initial velocity $=0,$ initlal acceleration $=0$
In general, acceleration can be represented as, $\frac{d a}{d t}=2$
On integration with respect to t and a, for lower limit $t=0$ to $t=t$ and for $a=0$ to $a=a$
$a=2 t$
Acceleration is rate of change of velocity,
$\frac{d v}{d t}=2 t$
$d v=2 t d t$
double integrating with respect to $\mathrm{t}$ and $\mathrm{v}$ for limits, $t=0$ to $t$ and $v=0$ to $v$
$\mathrm{v}=t^{2}$
$\frac{d x}{d t}=t^{2}$
Integrating for limits $t=0$ to $t=3$
$x=(3)^{2}-(0)^{2}=9 m $
Time $t=3 s$
Initial velocity $=0,$ initlal acceleration $=0$
In general, acceleration can be represented as, $\frac{d a}{d t}=2$
On integration with respect to t and a, for lower limit $t=0$ to $t=t$ and for $a=0$ to $a=a$
$a=2 t$
Acceleration is rate of change of velocity,
$\frac{d v}{d t}=2 t$
$d v=2 t d t$
double integrating with respect to $\mathrm{t}$ and $\mathrm{v}$ for limits, $t=0$ to $t$ and $v=0$ to $v$
$\mathrm{v}=t^{2}$
$\frac{d x}{d t}=t^{2}$
Integrating for limits $t=0$ to $t=3$
$x=(3)^{2}-(0)^{2}=9 m $
Standard 11
Physics
