(a) See graph in attachment
The appropriate graph to draw in this part is a graph of velocity vs time.
In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.
Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at
t = 10 s
v = 15 m/s
(b) [tex]1.5 m/s^2[/tex]
The average acceleration of the horse can be calculated as:
[tex]a=\frac{v-u}{t}[/tex]
where
v is the final velocity
u is the initial velocity
t is the time interval
In this problem,
u = 0
v = 15 m/s
t = 10 s
Substituting,
[tex]a=\frac{15-0}{10}=1.5 m/s^2[/tex]
(c) 75 m
For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s is the distance travelled
u is the initial velocity
t is the time interval
a is the acceleration
In this problem,
u = 0
t = 10 s
[tex]a=1.5 m/s^2[/tex]
Substituting,
[tex]s=0+\frac{1}{2}(1.5)(10)^2=75 m[/tex]
(d) See attached graphs
In a uniformly accelerated motion:
- The distance travelled (x) follows the equation mentioned in part c,
[tex]x=ut+\frac{1}{2}at^2[/tex]
So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.
- The acceleration is constant during the motion, and its value is
[tex]a=1.5 m/s^2[/tex] (calculated in part b)
therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.
