solid food containing starch is taken into the mouth at point A and enters the oesophagus at point B .State two differences between the state of the food at pointA and pointB​

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Acceleration - Motion

As per the question, the object starts from rest and it is attaining the speed of 50m/s. After that, the car is travelling a distance of 400m.

But according to the given information, the object starts from from rest, i.e, the initial velocity of the object will be 0, although the final velocity will have some value.

Here, we concluded that:

\implies \: u_{object} = 0 \: ms^{-1}⟹u

object

=0ms

−1

Simply applying the third equation of motion we can get the acceleration of the object. Third equation of motion deals with the final velocity, initial velocity, acceleration and distance travelled by the object respectively.

In mathematical term it would be like this;

\implies\: v^2 = u^2 + 2as⟹v

2

=u

2

+2as

Now simply applying the equation of kinematics :

\begin{gathered}\implies \: (50)^{2} = {(0)}^{2} + 2 \times a \times 400 \\ \\ \implies \: 2500= {(0)}^{2} + 2 \times a \times 400 \\ \\ \implies \: 2500 = 0 + 2 \times a \times 400 \\ \\ \implies \:2500 = 0 + 800 \times a \\ \\ \implies \:2500 = 0 + 800a \\ \\ \implies \:2500 = 800a \\ \\ \implies \:a = \cancel\frac{2500}{800} \\ \\ \implies \: \boxed{ \bf{a = 3.12 \: ms^{ - 2} }}\end{gathered}

⟹(50)

2

=(0)

2

+2×a×400

⟹2500=(0)

2

+2×a×400

⟹2500=0+2×a×400

⟹2500=0+800×a

⟹2500=0+800a

⟹2500=800a

⟹a=

800

2500

a=3.12ms

−2

So our final answer is :

Acceleration of the object is 3.12 m/s².

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