An automobile with an initial speed of 4.30 m/s accelerates uniformly at the rate of 3.00 m/s². find the final speed and the displacement after 5.00 s.

The final speed and the displacement of the automobile are 19.3 m/s and 59 m respectively

Step-by-step explanation:

From Newton's Equation of motion,

a = v-u/t Equation 1

making v the subject of the equation above,

v = at + u Equation 2

Where a = acceleration of the automobile, v = final speed, u = initial speed, t = time.

Given:a = 3.00 m/s², u = 4.3 m/s, t = 5.00 s

Substituting these values into equation 2,

v = 3(5) + 4.3

v = 15 + 4.3

v = 19.3 m/s.

Also using

v² = u² + 2as

Making s the subject of the equation,

s = (v²-u²)/2a

Where s = displacement.

Given: v= 19.3 m/s, u = 4.3 m/s, a = 3 m/s²

Substituting these values into equation 2

s = (19.3²-4.3²)/2×3

s = (372.49 - 18.49)/6

s = 354/6

s = 59 m

Thus the final speed and the displacement of the automobile are 19.3 m/s and 59 m respectively

definition: a circle is the  set of all points in a plane that are equidistant  from a given  point  called the center of the circle. we use the symbol ⊙ to represent a circle. the a line segment from the center of the circle to  any point  on the circle is a radius of the circle.

step-by-step explanation:

i googled it and this is what i got i don't know how it is. sorry.

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47