MHT CET · Maths · Differentiation
The equation of motion of a particle is \(s=a t^2+b t+c\). If the displacement after 1 second is 20 m , velocity after 2 seconds is \(30 \mathrm{~m} / \mathrm{sec}\) and the acceleration is \(10 \mathrm{~m} / \mathrm{sec}^2\), then
- A \(a+c=2 b\)
- B \(a+c=b\)
- C \(\mathrm{a}-\mathrm{c}=\mathrm{b}\)
- D \(a+c=3 b\)
Answer & Solution
Correct Answer
(B) \(a+c=b\)
Step-by-step Solution
Detailed explanation
\(s=a t^2+b t+c\)
Displacement after 1 second is 20 m .
\(\therefore \quad 20=a+b+c...(i)\)
\(\text { Velocity }=\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}=2 \mathrm{at}+\mathrm{b}\)
Velocity after 2 seconds is \(30 \mathrm{~m} / \mathrm{sec}\)
\(\therefore \quad 30=4 a+b...(ii)\)
Acceleration \(=\frac{\mathrm{dv}}{\mathrm{dt}}=2 \mathrm{a}\)
Acceleration is \(10 \mathrm{~m} / \mathrm{sec}^2\)
\(\therefore \quad 10=2 a\)
\(\Rightarrow a=5\)
From (ii), we get \(b=10\)
From (i), we get \(\mathrm{c}=5\)
\(\therefore \quad a+c=b\)
Displacement after 1 second is 20 m .
\(\therefore \quad 20=a+b+c...(i)\)
\(\text { Velocity }=\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}=2 \mathrm{at}+\mathrm{b}\)
Velocity after 2 seconds is \(30 \mathrm{~m} / \mathrm{sec}\)
\(\therefore \quad 30=4 a+b...(ii)\)
Acceleration \(=\frac{\mathrm{dv}}{\mathrm{dt}}=2 \mathrm{a}\)
Acceleration is \(10 \mathrm{~m} / \mathrm{sec}^2\)
\(\therefore \quad 10=2 a\)
\(\Rightarrow a=5\)
From (ii), we get \(b=10\)
From (i), we get \(\mathrm{c}=5\)
\(\therefore \quad a+c=b\)
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