A runner traveling with an initial velocity of 1.7 m/s accelerates at a constant rate of 1.2 m/s2 for a time of 6 seconds. What distance does the runner cover during this process?

31.80m

Explanation:

Given the following :

Initial Velocity (u) = 1.7m/s

Acceleration (a) = 1.2m/s²

Time (t) = 6seconds

Distance covered during the process:

Using the motion equation; calculate the final velocity (v)

v = u + at

v = 1.7 + (1.2*6)

v = 1.7 + 7.2

v = 8.9m/s

We can then find the distance covered during the process using :

v² = u² + 2aS

Where ; v = final velocity ; u = initial velocity

S = distance and a = acceleration

Hence,

8.9² = 1.7² + (2 × 1.2 × s)

8.9² = 2.89 + (2.4 × s)

79.21 = 2.89 + 2.4s

79.21 - 2.89 = 2.4s

76.32 = 2.4s

s = 76.32 / 2.4

s = 31.8m

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find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2