Question

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$\huge{\sf Figure{\uparrow}}$

A particle is moving along a straight line such that its velocity varies with position as shown in the Figure above ,Then the acceleration of the particle at x=10m/s?

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Answer 1

Answer:

-8.888 m/s

Explanation:

Using calculus for this question, we can say that

Acceleration = vdv/dx

Now first let's find out v,

So we need to know the equation of the line,

Using point slope form, let's take x coordinates as X and y coordinates as v

Formula says,

Y-y1 = m(x-x1)

= here replacing y by v

m = 0-20/15-0 = - 4/3

V-20= - 4/3(x-0)

V-20 = - 4/3 x

Multiplying throughout by 3,

3v-60= - 4x

3v+4x-60 = 0

3v+4x = 60

We need to find v at x = 10,

3v + 40 = 60

3v = 20

V = 20/3

Now let's find dv/dx

We have the equation v-20 = - 4/3x

V = - 4/3x + 20

Now differentiating wrt x

Dv/dx = - 4/3

Acceleration = vdv/dx

= 20/3((-4/3)

= - 80/9 m/s

= - 8.888 m/s