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

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6

you cant move and you feel like someones pushing you down onto your bed it can feel like you want to get up but you just can't it can be extremely scary