motion
along
particle
straight
described
by
the
equation
x= 6 + 27+ t3
Find the time when
K.E of the particle
is zero.
Answers
Answer:
The particle’s retardation is 12\ m/s^212 m/s
2
Given:
x=8+12 t-t^{3}x=8+12t−t
3
Solution:
We know that,
On differentiating, displacement, we get velocity.
\Rightarrow \text { Velocity }(v)=\frac{d x}{d t}=12-3 t^{2}⇒ Velocity (v)=
dt
dx
=12−3t
2
Again on differentiating, velocity, we get acceleration.
\Rightarrow \text { Acceleration }(a)=\frac{d v}{d t}=-6 t⇒ Acceleration (a)=
dt
dv
=−6t
When velocity becomes zero,
12-3 t^{2}=0 \Rightarrow 12=3 t^{2} \Rightarrow 4=t^{2} \Rightarrow t=2 s12−3t
2
=0⇒12=3t
2
⇒4=t
2
⇒t=2s
Therefore,
=-6 t=-6 \times 2=-12 \ \mathrm{m} / \mathrm{s}^{2}=−6t=−6×2=−12 m/s
2
In the above solution, the answer we got is negative, that’s because it is negative acceleration, also known as retardation.
To calculate the retardation, we took the approach of calculus and we used the concept that the acceleration is the differentiation of velocity which in turn is derived by differentiating the distance. By definition, acceleration is rate of change of velocity.
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