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Chapter 2: Problem 196

The distance travelled by a particle is given by $\mathrm{s}=3+2 \mathrm{t}+5 \mathrm{t}^{2}\( The initial velocity of the particle is \)\ldots$ (A) 2 unit (B) 3 unit (C) 10 unit (D) 5 unit

Short Answer

Expert verified
The initial velocity of the particle is \(s'(0) = 2 + 10(0) = 2\), which is (A) 2 units.

Step by step solution

01

Identify the given function

The distance travelled by the particle as a function of time is given by: s(t) = 3 + 2t + 5t^2
02

Find the first derivative of s(t) with respect to t

Using the power rule, the first derivative of s(t) with respect to t is: s'(t) = \(\frac{d}{dt}(3) + \frac{d}{dt}(2t) + \frac{d}{dt}(5t^2)\)
03

Apply the power rule for each term

The power rule states that \(\frac{d}{dt}(k*t^n)=k*n*t^{n-1}\), where k and n are constants. Applying this rule to each term, we have: s'(t) = \(\frac{d}{dt}(3) + \frac{d}{dt}(2t) + \frac{d}{dt}(5t^2)\) = 0 + 2 + 10t
04

Evaluate s'(t) at t = 0

The initial velocity corresponds to the value of s'(t) when t = 0: s'(0) = 2 + 10(0) = 2
05

Identify the correct option

Comparing the result with the given options, the correct answer is (A) 2 units. So, the initial velocity of the particle is 2 units.

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