Question

The motion of a particle along a straight line is described by equations x=8+12t-t^3,where x is in metre and t in second.The retardation of the particle when it's velocity becomes zero is

Answer 1

Hey There!!

Here,we are given:

$x=8+12t-t^3$


Let us find the expression of velocity:

$v = \frac{dx}{dt} = 12 - 3t^2$

We need the value of acceleration when velocity becomes zero. For this, we need to find the time when velocity becomes zero.

$12-3t^2=0 \\ \\ \implies t^2=4 \\ \\ \implies t=2 \, \,s$


Also, let us find the acceleration :

$a = \frac{dv}{dt} = -6t$

Now, velocity becomes zero at t = 2 s. So, we need to find value of acceleration at t = 2

We have:

$a = -6t \\ \\ \implies a = -6 \times 2 \\ \\ \implies \boxed{a=-12 \, \, m / s^2}$


Thus, the retardation is $12 \, \, m / s^2$


Hope it helps
Purva
Brainly Community

Answer 2

$\text{displacement,} \: \sf\:x = 8 + 12t - t^3 \\ \\ \text{Velocity,}\ v = \frac{dx}{dt}= \frac{d}{dt} \big(8+12t-t^3 \big) = 12-3t^2\\ \\ \text{acceleration,}\ a= \frac{dv}{dt}= \frac{d}{dt} \big(12-3t^2 \big)=-6t\\ \\ \textsf{When velocity becomes 0}\\ \\ \Rightarrow v=0\\ \\ \Rightarrow\sf\: 12-3t^2=0\\ \\ \Rightarrow 3t^2=12\\ \\ \Rightarrow t^2=\frac{12}{3}=4\\ \\ \Rightarrow t=\sqrt{4}=2\ seconds\\ \\ \texttt{acceleration at that time is given as}\\ \\ a_{(t=2)}=-6 \times 2 = -12\ ms^{-2}\\ \\ \texttt{Required retardation is 12 ms {}^{-2} }$