# The displacement function of a particle is f(x) = ax^n + bx + 4. What should be the value of n for the particle to have a constant acceleration.

Asked on by lxsptter

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The dispalcement function of the particle is given by f(x) = ax^n + bx + 4. Here, x has to represent time; f(x) is a function of time. It would be more appropriate to write this as f(t) = a*t^n + b*t + 4

The first derivative of f(t) gives the velocity of the particle, v(t) = f'(t) = a*n*t^(n - 1) + b

The second derivative of f(t) gives the acceleration of the particle, a = f''(t) = a*n*(n - 1)*t^(n - 2)

For a constant acceleration, f''(t) = a*n*(n - 1)*t^(n - 2) should be a constant.

This is the case when n - 2 = 0

=> n = 2

The particle has constant acceleration if n = 2 in f(t) = at^n + bt + 4.

quantatanu | Student, Undergraduate | (Level 1) Valedictorian

Posted on

Displacement f(x) = ax^n + bx + 4. where x is time. Now acceleration is the second derivative of displacement, that is " a = f''(x) ", if acceleration has to be constant that means derivative of acceleration "a" must be "0", that means (d/dx) a = f'''(x) = 0,

now

f'(x) = a*n*x^(n - 1) + b

=> f''(x) = a*n*(n - 1)*x^(n - 2)

=> f'''(x) = a*n*(n-1)*(n-2)*x^(n-3)

we need,  f'''(x) = 0, and this is possible in three ways,

n = 0, 1, 2

so for n= 0 or 1 or 2, the acceleration will be constant.

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