1. A particle moves along a straight line with its position at time t given by s(t) = 9t^3 - 3t^2 + 2t + 5. Find the velocity at time t. (i) Determine the velocity at time t = 2 se... 1. A particle moves along a straight line with its position at time t given by s(t) = 9t^3 - 3t^2 + 2t + 5. Find the velocity at time t. (i) Determine the velocity at time t = 2 sec. (ii) Determine the time velocity when the body is at rest. 2. The velocity of an object is given by v(t) = 6t^2 - 4t + 3. Find the acceleration as a function of time, and determine the acceleration at time t = 3 sec. 3. The position of an object is given by s(t) = t^4 - 8t + 1. Find the velocity and acceleration, and determine the acceleration at t = 0.5 sec. 4. The velocity of a car is given by v(t) = 20t - 4.9t^2. Find the acceleration at time t = 4 sec. Read more

Steps to Solve

1. Finding Velocity from Position Function

To find the velocity function, we differentiate the position function $s(t)$ with respect to $t$.

Given $s(t) = 9t^3 - 3t^2 + 2t + 5$, we find the derivative:

$$ v(t) = \frac{ds}{dt} = \frac{d}{dt}(9t^3 - 3t^2 + 2t + 5) = 27t^2 - 6t + 2 $$

2. Determine Velocity at t = 2 sec

Now, substitute $t = 2$ into the velocity function:

$$ v(2) = 27(2^2) - 6(2) + 2 = 27(4) - 12 + 2 = 108 - 12 + 2 = 98 $$

So, the velocity at time $t = 2$ seconds is $98$ units.

3. Finding When the Particle is at Rest

To determine when the particle is at rest, we set the velocity function equal to zero:

$$ 27t^2 - 6t + 2 = 0 $$

We can use the quadratic formula:

$$ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Where $a = 27$, $b = -6$, and $c = 2$.

4. Quadratic Formula Application

Calculating the discriminant:

$$ \Delta = (-6)^2 - 4(27)(2) = 36 - 216 = -180 $$

Since the discriminant is negative, there are no real roots, meaning the particle never comes to rest.

5. Finding Acceleration from Velocity Function

The acceleration function is found by differentiating the velocity function $v(t) = 6t^2 - 4t + 3$:

$$ a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 4t + 3) = 12t - 4 $$

6. Determine Acceleration at t = 3 sec

Substitute $t = 3$ into the acceleration function:

$$ a(3) = 12(3) - 4 = 36 - 4 = 32 $$

So, the acceleration at time $t = 3$ seconds is $32$ units.

7. Finding Velocity and Acceleration for the Third Function

For $s(t) = t^4 - 8t + 1$, we differentiate to find velocity:

$$ v(t) = \frac{ds}{dt} = 4t^3 - 8 $$

8. Finding Acceleration for the Third Function

Then, differentiate again for acceleration:

$$ a(t) = \frac{dv}{dt} = 12t^2 $$

9. Determine Acceleration at t = 0.5 sec

Substituting $t = 0.5$:

$$ a(0.5) = 12(0.5)^2 = 12(0.25) = 3 $$

So, the acceleration at time $t = 0.5$ seconds is $3$ units.

10. Finding Acceleration for the Fourth Function

For the velocity function $v(t) = 20t - 4.9t^2$, the acceleration function is:

$$ a(t) = \frac{dv}{dt} = 20 - 9.8t $$

11. Determine Acceleration at t = 4 sec

Substituting $t = 4$:

$$ a(4) = 20 - 9.8(4) = 20 - 39.2 = -19.2 $$

So, the acceleration at time $t = 4$ seconds is $-19.2$ units.