Question

A particle moves along a straight line with the equation of motion s = t^{5} - 2 t^{4} Find the value of t (other than 0 ) at which the acceleration is equal to zero.

Answer 1

Answer:

t = 8/5

Step-by-step explanation:

I'm assuming the equation given is the velocity of the particle

Derive the equation

$5t^4 - 8t^3$

Set it equal to 0 & solve for t

$0 = 5t^4 - 8t^3\\8t^3 = 5t^4\\8 = 5t\\\frac{8}{5} = t$

Answer 2

The value of time at which acceleration is zero is required.

The required value of $t=1.2\ \text{s}$

The function of motion is

$s=(t)^5-2t^4$

Differentiating with time we get velocity

$s'=5t^4-8t^3$

Again differentiating with time we get acceleration

$s''=20t^3-24t^2$

Now acceleration $s''=0$

$0=20t^3-24t^2\\\Rightarrow 20t^3=24t^2\\\Rightarrow t=\dfrac{24}{20}\\\Rightarrow t=1.2\ \text{s}$

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