Speeds of a particle at 3rd and 8th sec are 20 ~m / s and zero respectively, then average acceleration between 3 rd and 8th sec will be

Correct answer: (B) 4 ~m / s^2. Time interval between 8 th sec. and 3rd sec Δ t=8-3=5 ~s i.e. Δ t=5 ~s Change in velocity Δ V=20-0=20 ~m / s ∴ Average acceleration =(Δ V)/(Δ t)=20/5=4 ~m / s^2

Q105 / 180 · JIPMER 2019 · Physics

Speeds of a particle at 3rd and 8th sec are $20 \mathrm{~m} / \mathrm{s}$ and zero respectively, then average acceleration between 3 rd and 8th sec will be

A
$3 \mathrm{~m} / \mathrm{s}^2$
Correct answer
B
$4 \mathrm{~m} / \mathrm{s}^2$
C
$5 \mathrm{~m} / \mathrm{s}^2$
D
$6 \mathrm{~m} / \mathrm{s}^2$

Solution

Time interval between 8 th sec. and 3rd sec $\Delta \mathrm{t}=8-3=5 \mathrm{~s}$ i.e. $\Delta t=5 \mathrm{~s}$ Change in velocity $\Delta \mathrm{V}=20-0=20 \mathrm{~m} / \mathrm{s}$ $\therefore$ Average acceleration $=\frac{\Delta \mathrm{V}}{\Delta \mathrm{t}}=\frac{20}{5}=4 \mathrm{~m} / \mathrm{s}^2$

Hence, the correct answer is option: 'B'.

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