Question

The height of an object thrown upward from the surface of a planet is given by the function s(t) = -2t2 + 28t, where t is the time in seconds and s(t) is the height in meters. What is the speed of the object at time t = 4?

Answer 1

Answer:

The speed of the object at time, t=4 is $20ms^{-1}$

Step-by-step explanation:

The speed is calculated using the formula;

$Speed=\frac{Distance}{Time\:taken}$

The distance covered by the object is given in terms of t, as

$s(t)=-2t^2+28t$

At time t=4, the distance covered by the object is;

$s(4)=-2(4)^2+28(4)$

$s(4)=80m$

The speed can now be calculated by substituting t=4 and distance 80 meters to obtain;

$Speed=\frac{80}{4} =20ms^{-1}$

Answer 2

Answer:

12 m/sec.

Step-by-step explanation:

The height of an object thrown upward from the surface of a planet is given by the function s(t) = - 2t² + 28t.

Here t = time

and s(t) = height in time t

Since the object has been thrown upwards under gravity so speed of ball will decrease with the time.

That means it we calculate rate of change in speed, $(\frac{ds}{dt})$

We can get the velocity at any moment of time.

So we will differentiate or find the derivative of the given expression.

s(t) = -2t² + 28t

$\frac{d[s(t)]}{dt}=\frac{d[-2t+28t^{2}] }{dt}$

V(t) = (-4t +28 )

Now at t = 4 velocity will be

V (4) = -4 × 4 +28 = -16 +28 = 12 m/sec.

The speed of the object is 12 meter per second.