Question 106: Motion in 1D

Question

The displacement and the increase in the velocity of a moving particle in the time interval of $$t$$ to $$(t+1) \mathrm{s}$$ are $$125 \mathrm{~m}$$ and $$50 \mathrm{~m} / \mathrm{s}$$, respectively. The distance travelled by the particle in $$(\mathrm{t}+2)^{\mathrm{th}} \mathrm{s}$$ is _________ m.

Accepted Answer

The displacement and the increase in the velocity of a moving particle from time $$t$$ to $$(t + 1) \mathrm{s}$$ are $$125 \mathrm{~m}$$ and $$50 \mathrm{~m} / \mathrm{s}$$, respectively. The distance traveled by the particle in $$(\mathrm{t} + 2)^{\mathrm{th}} \mathrm{s}$$ is calculated as follows:
Given that the acceleration is constant, we start with:
$$ v = u + at $$
When the velocity has increased by $$50 \mathrm{~m}/\mathrm{s}$$, the equation becomes:
$$ u + 50 = u + a \quad \Rightarrow \quad a = 50 \mathrm{~m}/\mathrm{s}^2 $$
Next, we consider the displacement:
$$ 125 = u t + \frac{1}{2} a t^2 $$
Since this is given over a unit time interval (from $$t$$ to $$(t + 1)$$), we use:
$$ 125 = u + \frac{a}{2} $$
Substituting $$a = 50 \mathrm{~m}/\mathrm{s}^2$$:
$$ 125 = u + \frac{50}{2} \quad \Rightarrow \quad  125 = u + 25 \quad \Rightarrow \quad u = 100 \mathrm{~m}/\mathrm{s} $$
To find the distance traveled by the particle in $$(t + 2)^	ext{th}$$ second, we use:
$$ S_n = u + \frac{a}{2} [2n - 1] $$
For $$n = t + 2$$ (i.e., the (t+2)th second):
$$ S_{(t+2)} = u + \frac{a}{2} [2(t+2) - 1] $$
With $$u = 100$$ and $$a = 50$$:
$$ S_{(t+2)} = 100 + \frac{50}{2} [2(t+2) - 1] = 100 + 25 	imes [2(t+2) - 1] = 100 + 25 	imes (2t + 4 - 1) = 100 + 25 	imes (2t + 3) = 100 + 25 	imes 5 = 100 + 125 = 225 \mathrm{~m} $$

Motion in 1D

The displacement and the increase in the velocity of a moving particle in the time interval of $t to (t+1) \mathrm{s} are 125 \mathrm{~m} and 50 \mathrm{~m} / \mathrm{s}, respectively. The distance travelled by the particle in (\mathrm{t}+2)^{\mathrm{th}} \mathrm{s}$ is _________ m.

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