Question

A particle moves along the x-axis so that its position at time 1 > 0 is given by x(t) and dx/dt = -10t⁴+9t² + 8t. The acceleration of the particle is zero when t=
a. 0.387
b. 0.831
c. 1.243
d. 1.647
e. 8.094​

Answer 1

Given:

A particle moves along the x-axis so that its position at time t > 0 is given by x(t).

$\dfrac{dx}{dt}=-10t^4+9t^2+8t$

To find:

The value of t at which acceleration of the particle is zero.

Solution:

We have, x(t) as position function. So, its derivative with respect to t is velocity.

$v=\dfrac{dx}{dt}=-10t^4+9t^2+8t$

Derivative of velocity with respect to t is acceleration.

$a=\dfrac{dv}{dt}=\dfrac{d}{dt}(-10t^4+9t^2+8t)$

$a=-10(4t^3)+9(2t)+8(1)$

$a=-40t^3+18t+8$

Putting a=0, to find the time t at which acceleration of the particle is zero.

$0=-40t^3+18t+8$

$0=-40t^3+18t+8$

Using the graphing calculator, we get t=0.831 is the only real solution of this equation.

$t=0.831$

Therefore, the correct option is b.