an athlete sprints from 150 m south of the finish line to 65 m south of the finish line in 5.0s what is his average velocity

Physics

2 answers:

MatroZZZ [7]2 years ago

6 0

Total displacement=150-65m=85m
Time=5s

$\\ \rm\longrightarrow Avg\:Velocity=\dfrac{Total\:Displacement}{Total\:Time}$

$\\ \rm\longrightarrow Avg\:Velocity=\dfrac{85}{5}$

$\\ \rm\longrightarrow Avg\:Velocity= 17m/s$

ladessa [460]2 years ago

5 0

Answer:

$displacement = 150 - 65 = 85m \: \\ time = 5sec \\ velocity = displacement/ time= \frac{85}{5} = 17 \frac{m}{s} \: south\\ thank \: you$

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<u>Explanation:</u>

Given data:

Mass of the box = 3.74 kg

Flat friction-less ground is pulled forward by a 4.20 N force at a 50.0 degree angle and pulled back by a 2.25 N force at a 122 degree angle.

First, we need to find the net horizontal force acting on the box. With the given data, the equation can be formed as below. Net horizontal force acting on the box (F) is given by

$F=\left(4.20 \times \cos 50^{\circ}\right)+\left(2.25 \times \cos 122^{\circ}\right)$

$F=(4.20 \times 0.64278)+(2.25 \times-0.5299)$

F = 2.699676 – 1.192275 = 1.507 N

Next, find acceleration of the box using Newton's second law of motion. This states that the link between mass (m) of an objects and the force (F) required to accelerate it. The equation can be given as

$F=m \times acceleration\ (a)$