Consider a particle initially moving with a velocity of $5 \mathrm{~m} \mathrm{~s}^{-1}$ starts decelerating at a constant rate of $2 \mathrm{~m} \mathrm{~s}^{-2}$.
(a) Determine the time at which the particle becomes stationary.
(b) Find the distance travelled in the second second.
(c) Find the distance travelled in the third second.

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Aicharya · Accepted Answer

(a) Here $u=5 \mathrm{~m} \mathrm{~s}^{-1}, a=-2 \mathrm{~m} \mathrm{~s}^{-2}, v=0, t=?$ Using $v=u+a t$, where $0=5-2 t \Rightarrow t=2.5 \mathrm{~s}$ $u=5 \mathrm{~m} \mathrm{~s}^{-1}, a=-2 \mathrm{~m} \mathrm{~s}^{-2}, n \leq 2$ $(b)$ Here Using $x_{n}=u+\frac{a}{2}(2 n-1)=5-\frac{2}{2}(2(2)-1)=2$ m (c) Here, if we use the above formula, we will get $x_{n}=0$, but in reality it is not zero. This formula is not applicable for the third second because velocity becomes zero in the third second, i.e., at $t=2.5 \mathrm{~s}$. The particle has a turning point at $t=2.5 \mathrm{~s}$. We have to indirectly calculate the distance travelled in this particular second. That is, we have to determine the distance travelled between $2 \mathrm{~s} \leq t \leq 2.5 \mathrm{~s}$ and $2.5 \mathrm{~s} \leq t \leq 3 \mathrm{~s}$, and then add the two. Displacement of the particle at $t=2.5 \mathrm{~s}$ is $x_{2.5}=\frac{u^{2}}{2 a}=\frac{(5)^{2}}{2(2)}=6.25 \mathrm{~m}$ Due to symmetry, the displacement of the particle at $t=2 \mathrm{~s}$ and $t=3 \mathrm{~s}$ are same, i.e., $x_{3}=x_{2}=5(2)+\frac{1}{2}(-2)(2)^{2}=6$ Thus, the distance travelled in the third second is $\begin{aligned} x &= (x_{2.5}-x_{2})+(x_{2.5}-x_{3}) \ &= (6.25-6)+(6.25-6)=0.5 \text{~m} \end{aligned}$

Consider a particle initially moving with a velocity of \(5 \mathrm{~m} \mathrm{~s}^{-1}\) starts…

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