Question

The height of a projectile t seconds after it is launched straight up in the air is given by f (t )equals negative 16 t squared plus 238 t plus 3. Find its acceleration at 5 seconds.

Answer 1

Answer:

$\displaystyle a(5)=-32$

Explanation:

Instant Acceleration

The kinetic magnitudes are usually related as scalar or vector equations. By doing so, we are assuming the acceleration is constant over time. But when the acceleration is variable, the relations are in the form of calculus equations, specifically using derivatives and/or integrals.

Let f(t) be the distance traveled by an object as a function of the time t. The instant speed v(t) is defined as:

$\displaystyle v(t)=\frac{df}{dt}$

And the acceleration is

$\displaystyle a(t)=\frac{dv}{dt}$

Or equivalently

$\displaystyle a(t)=\frac{d^2f}{d^2t}$

The given height of a projectile is

$f(t)=-16t^2 +238t+3$

Let's compute the speed

$\displaystyle v(t)=-32t+238$

And the acceleration

$\displaystyle a(t)=-32$

It's a constant value regardless of the time t, thus

$\boxed{\displaystyle a(5)=-32}$