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Chapter 2: Problem 227

A particle is moving in a straight line with initial velocity of $200 \mathrm{~ms}^{-1}\( acceleration of the particle is given by \)\mathrm{a}=3 \mathrm{t}^{2}-2 \mathrm{t}$. Find velocity of the particle at 10 second. (A) \(1100 \mathrm{~ms}^{-1}\) (B) \(300 \mathrm{~ms}^{-1}\) (C) \(900 \mathrm{~ms}^{-1}\) (D) \(100 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The velocity of the particle at 10 seconds is (A) \(1100\mathrm{~ms}^{-1}\).

Step by step solution

01

Given Information

Initial Velocity: \(v_0=200\mathrm{~ms}^{-1}\) Acceleration function: \(a(t)=3t^2-2t\) We use the following formula to find the velocity function, where \(C\) represents the integration constant: \[v(t)=\int a(t)\, dt + C\] #Step 2: Integrate the acceleration function#
02

Integration of Acceleration Function

Integrate the function \(a(t)\) with respect to \(t\): \[v(t)=\int(3t^2-2t) \, dt = t^3 - t^2+C\] #Step 3: Determine the integration constant#
03

Find Integration Constant

We know that the initial velocity is \(200 \mathrm{~ms}^{-1}\) at \(t=0\), we have: \[v(0)=0^3 - 0^2+C=200\] So, the integration constant \(C=200\). #Step 4: Write down the velocity function#
04

Velocity Function

Now that we have found the integration constant, we can write down the velocity function: \[v(t)=t^3 - t^2+200\] #Step 5: Find the velocity at 10 seconds#
05

Velocity at 10 Seconds

Plug in \(t=10\) into the velocity function to find the velocity at 10 seconds: \[v(10)=(10)^3 - (10)^2+200=1000-100+200=1100\mathrm{~ms}^{-1}\] The velocity of the particle at \(10\) seconds is \(1100\mathrm{~ms}^{-1}\). The correct answer is (A) \(1100\mathrm{~ms}^{-1}\).

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