Respuesta :
we know that
The scalar magnitude of the velocity vector is the speed. The speed is equal to
[tex]Speed=\frac{distance}{time}[/tex]
in this problem we have
[tex]distance=400\ m \\time=120\ sec[/tex]
substitute in the formula
[tex]Speed=\frac{400}{120}[/tex]
[tex]Speed=3.5\frac{m}{sec}[/tex]
therefore
the answer Part a) is
the speed is equal to [tex]3.5\frac{m}{sec}[/tex]
Part b) Find the velocity
we know that
Velocity is a vector quantity; both magnitude and direction are needed to define it
in this problem we have
the magnitude is equal to the speed
[tex]magnitude=3.5\frac{m}{sec}[/tex]
[tex]direction=North\ East\ (NE)[/tex]
therefore
the answer Part b) is
the velocity is [tex]3.5\frac{m}{sec}\ North\ East\ (NE)[/tex]
Part c)
we know that
the acceleration is equal to the formula
[tex]a=\frac{V2-V1}{t2-t1}[/tex]
in this problem we have
[tex]V2=0 \\V1=3.5\frac{m}{sec}[/tex]
[tex]t2=15\ sec\\t1=0[/tex]
substitute in the formula
[tex]a=\frac{0-3.5}{15-0}[/tex]
[tex]a=-\frac{3.5}{15}\frac{m}{sec^{2}}[/tex]
[tex]a=-\frac{7}{30}\frac{m}{sec^{2}}[/tex]
therefore
the answer Part c) is
the acceleration is [tex]-\frac{7}{30}\frac{m}{sec^{2}}[/tex]
This is an example of negative acceleration
Step-by-step explanation:
1. It is given that,
Distance covered by the cyclist, d = 400 m
Time taken, t = 120 s
Speed = distance / time taken
[tex]v=\dfrac{400\ m}{120\ s}[/tex]
v = 3.33 m/s
So, the speed of the cyclist is 3.33 m/s.
2. The cyclist comes to a stop position within 15 seconds. Its acceleration is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{0-3.33\ m/s}{15\ s}[/tex]
[tex]a=-0.22\ m/s^2[/tex]
So, the acceleration of the cyclist is [tex]-0.22\ m/s^2[/tex]. It is an example of negative acceleration as the cyclist is decelerating. Hence, this is the required solution.
