Question

Suppose the position of an object moving horizontally after t seconds is given by the following functions s=f(t), s equals f , open t close , comma where s is measured in feet, with s>0 s greater than 0 corresponding to positions right of the origin.

(a) Graph the position function.
(b) Find and graph the velocity function. When is the object stationary, moving to the right, and moving to the left?
(c) Determine the velocity and acceleration of the object at t=1. t equals 1 .
(d) Determine the acceleration of the object when its velocity is zero.
(e) On what intervals is the speed increasing?

f(t)= -t^2 +4t-3;0<=t<=5

Answer 1

Answer:

(a) The graph of position function is shown below.

(b) The velocity function is $v(t)=-2t+4$ and the graph of position function is shown below. The object is stationary at t=2, moving to the right at t>2, and moving to the left at t<2.

(c) Velocity and acceleration of the object at t=1 are 2 and -2 respectively.

(d) The acceleration of the object is -2 when its velocity is zero.

(e) The speed is not increasing at any interval because the acceleration is constant.

Step-by-step explanation:

(a)

The given function is

$f(t)=-t^2+4t-3; 0\leq t\leq 5$

The position of an object moving horizontally after t seconds is given by

$s=f(t)=-t^2+4t-3$

The graph of position function is shown below.

(b)

Differentiate the position function with respect to time to find the velocity function.

$v=f'(t)=-2t+4$

Put v=0 to find the time when the object is stationary.

$0=-2t+4$

$2t=4$

$t=2$

The object is stationary at t=2 because the velocity of the object is 0 at t=2.

The velocity function is $v(t)=-2t+4$ and the graph of position function is shown below. The object is stationary at t=2, moving to the right at t>2, and moving to the left at t<2.

(c)

The velocity function is

$v=f'(t)=-2t+4$

Substitute t=1 in the above function.

$v=f'(1)=-2(1)+4=2$

Differentiate the velocity function with respect to time to find the acceleration function.

$a=f''(t)=-2$

Substitute t=1 in the above function.

$a=f''(1)=-2$

Therefore the velocity and acceleration of the object at t=1 are 2 and -2 respectively.

(d)

The velocity of the object is 0 at t=2.

Substitute t=2 in the acceleration function to find the acceleration of the object when its velocity is zero.

$a=f''(2)=-2$

The acceleration of the object is -2 when its velocity is zero.

(e)

The acceleration function of the object is

$a=f''(t)=-2$

It is a constant function.

Therefore the speed is not increasing at any interval because the acceleration is constant.