Question

The position of a ball Rolling in a straight line is given by X equal to 2 + 6.6 t minus 1.1 t square what is the velocity at 1. t=2s and 2. t= 3s

Answer 1

Answer:

1. 2.2 units/s

2. 0 units/s

Explanation:

Given:

Position of a ball rolling in a straight line is given by:

$X = 2+6.6t-1.1t^{2}$

To find:

Velocity at

1. t = 2s

2. t = 3s

Solution:

First of all, let us have a look a the formula of Velocity, V:

$V = \dfrac{D}{T}$

Where D is displacement and

T is the time taken.

We are given that:

$X = 2+6.6t-1.1t^{2}$

If we differentiate X w.r. to t, we will get Velocity as per above formula.

Differentiating X w.r. to t:

$\dfrac{d}{dt}X = \dfrac{d}{dt}(2+6.6t-1.1t^{2})\\\Rightarrow V=0 + 6.6 \times 1 \times t^{1-1} - 2 \times 1.1 \times t^{2-1}\\\Rightarrow V=6.6 \times t^{0} - 2 \times 1.1 \times t\\\Rightarrow V=6.6 - 2.2 \times t$

Formula used:

$1.\ \dfrac{d}{dt} (Constant) = 0\\2.\ \dfrac{d}{dt} (t^n) = n\times t^{n-1}$

Now, solving the questions:

1. Putting t = 2s in $V=6.6 - 2.2 \times t$

$\Rightarrow V =6.6 - 2.2 \times 2\\\Rightarrow V =6.6 - 4.4\\\Rightarrow V =2.2\ units/s$

2. Putting t = 2s in $V=6.6 - 2.2 \times t$

$\Rightarrow V =6.6 - 2.2 \times 3\\\Rightarrow V =6.6 - 6.6\\\Rightarrow V =0\ units/s$