Velocity and Acceleration

**Dean** · August 15th 2007, 10:08 PM

A body moves x metres in t seconds, where $x = t^3-7t^2+3<br />$

Find:

{i} the velocity at t = 6 seconds
{ii} the acceleration at t = 8 seconds.

**earboth** · August 15th 2007, 10:28 PM

Originally Posted by Dean

A body moves x metres in t seconds, where $x = t^3-7t^2+3<br />$

Find:

{i} the velocity at t = 6 seconds
{ii} the acceleration at t = 8 seconds.

Hello,

1. I assume that you are modeling a continuously(?) accelerated motion. (I'm pretty sure that this expresion is not accurate but I don't have a special (Physics) dictionary)

2. The value of x is the distance after t seconds of movement:

$d(t) = t^3-7t^2+3$

3. The speed after t seconds is:

$v(t) = d'(t) = 3t^2-14t~\Longrightarrow~ v(6) = 24\frac{m}{s}$

4. The acceleration is:

$a(t) = v'(t) = d''(t) = 6t-14~\Longrightarrow~ a(8) = 34\frac{m}{s^2}$

**iced_maggot** · August 15th 2007, 10:36 PM

absolutely right, just remember that

Displacement ----> Velocity(Displacement/Time) ----> Acceleration (Change of Speed/Time) , Where the arrows indicates a derivative (e.g. the derivative of Displacement gives Velocity)

If you need to go the other way, just integrate, remembering to add c, which you will have to work out from the info given by the question

**topsquark** · August 16th 2007, 05:57 AM

Originally Posted by Dean

A body moves x metres in t seconds, where $x = t^3-7t^2+3<br />$

Originally Posted by earboth

Hello,

1. I assume that you are modeling a continuously(?) accelerated motion. (I'm pretty sure that this expresion is not accurate but I don't have a special (Physics) dictionary)

For the record, this is motion under a constant "jerk." Just as $v = \frac{dx}{dt}$ and $a = \frac{dv}{dt}$ , $j = \frac{da}{dt}$ .

-Dan

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