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

the focus of lesson 1 is newton's first law of motion - sometimes referred to as the law of inertia. newton's first law of motion is often stated as. an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

explanation:

the device is called “ rocket engine”