Starting from rest the motor boat travels around the circular path (radius of curvature= 40 m),at a speed (v= 0.8 t m/s),where (t) is in seconds. The magnitude of the boat's velocity when it has traveled 20 m is

The magnitude of the boat's velocity is 5.66 m/s

Step-by-step explanation:

Given;

Radius of curvature of the circular path, r = 40 m

speed of the boat, v = 0.8 t m/s

The magnitude of the boat's velocity when it has traveled 20 m is ?

a = dv / dt,

v = 0.8 t (differentiate with respect to t)

dv / dt = 0.8 = a

thus, a = 0.8 m/s²

If the boat starts from rest, initial velocity, u = 0

Apply kinematic equation, to solve for the boat velocity at 20 m

v² = u² + 2as

v² = 0 + 2 x 0.8 x 20

v² = 0 + 32

v² = 32

v = √32

v = 5.66 m/s

Therefore, the magnitude of the boat's velocity when it has traveled 20 m is 5.66 m/s

a.   we have two

lines:   y = 4-x   and  

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b.   4-x = 8-x^-1

  x           4-x       8-x^-1

-3           7        

8.33

-2           6         8.5

-1           5         9

  0           4        

-

  1           3        

7

  2           2        

7.5

  3           1         7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:  

x=-4.236

y = 4-x   = 4 – (-4.236)

= 8.236

y = 8-x^-1 =   .236)^-1

= 8.236

x=0.236

  y = 4-x   = 4 – (0.236) = 3.764

y = 8-x^-1 =   8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c.   to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x   and   y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines.   the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!