1) The acceleration of the runner is [tex]-3.33 m/s^2[/tex]
2) The time taken is 4.38 h
3) The final velocity is 13.7 m/s north
4) Displacement: 10 m forward, distance: 40 m
Explanation:
1)
The acceleration of a body is given by the equation
[tex]a=\frac{v-u}{t}[/tex]
where
v is the final velocity
u is the initial velocity
t is the time taken for the velocity to change from u to v
For the runner in this problem, we have
u = 10 m/s
v = 0 (he comes to a stop)
t = 3 s
Substituting, we find
[tex]a=\frac{0-10}{3}=-3.33 m/s^2[/tex]
2)
The speed of an object in uniform motion is given by
[tex]v=\frac{d}{t}[/tex]
where
d is the distance covered
t is the time taken
For the car in this problem, we have:
d = 350 km (distance covered)
v = 80 km/h (speed)
Solving for t, we find the time required for the trip:
[tex]t=\frac{d}{v}=\frac{350}{80}=4.38 h[/tex]
3)
For an object in uniformly accelerated motion, we can use the following suvat equation:
[tex]v=u+at[/tex]
where
u is the initial velocity
v is the final velocity
a is the acceleration
t is the time
For the object in this problem, taking north as positive direction, we have
u = 5.0 m/s (initial velocity)
[tex]a=3.0 m/s^2[/tex] (acceleration)
And substituting t = 2.9 s, we find the final velocity:
[tex]v=5.0 + (3.0)(2.9)=13.7 m/s[/tex] (north)
4)
a) Displacement is a vector connecting the initial position to the final position of motion. Its magnitude is equal to the shortest distance in a straight line between the initial and the final position.
In this problem, you walk 25 meters forward and then 15 meters backward: therefore, the final position is
x = 25 - 15 = +10 m (forward)
Therefore, the displacement is
[tex]\Delta x = +10 - 0 = +10 m[/tex] (forward)
b) Distance is a scalar quantity measuring the total length of the path covered during the motion, regardless of the direction. In this case therefore, distance is simply calculated by adding the distance covered in the first and second part of the motion:
[tex]d=25 + 15 = 40 m[/tex]
Learn more about velocity, acceleration, distance and displacement:
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