The Jamaican Bobsled Team is sliding down a hill in a toboggan at a rate of 5 m/s when he reaches an even steeper slope. If he accelerates at 2 m/s2 for the 5 m slope, how fast is he traveling when he reaches the bottom of the 5 m slope?

6.71 m/s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 5 m/s

Acceleration (a) = 2 m/s²

Distance (s) = 5 m

Final velocity (v) =.?

The final velocity at the bottom of the slope can be obtained as shown below:

v² = u² + 2as

v² = 5² + (2 × 2 ×5)

v² = 25 + 20

v² = 45

Take the square root of both side

v = √45

v = 6.71 m/s

Thus, the velocity of the Jamaican Bobsled Team when he reached the bottom of the slope is 6.71 m/s

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6