Could you help answer all these questions please I need it ASAP Its the PDF Pretty please

Suppose the position of an object moving horizontally after t seconds is given by the following functions s=f(t), s equals f , o

Scilla [17]

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.

7 0

3 years ago

Can somebody please help me with this

Paul [167]

∠PYG = 137°

Step-by-step explanation:

We know that a straight line is equal to 180°, so by looking at the image we can tell that if we added both of these angles together they would be equal to the straight line.

We can make an equation by adding these two angles together and setting them equal to 180.

$2p+153+6p+91=180$