(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) ![1.5 m/s^2](https://tex.z-dn.net/?f=1.5%20m%2Fs%5E2)
The average acceleration of the horse can be calculated as:
![a=\frac{v-u}{t}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv-u%7D%7Bt%7D)
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,
![a=\frac{15-0}{10}=1.5 m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B15-0%7D%7B10%7D%3D1.5%20m%2Fs%5E2)
(c) 75 m
For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:
![s=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
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
![a=1.5 m/s^2](https://tex.z-dn.net/?f=a%3D1.5%20m%2Fs%5E2)
Substituting,
![s=0+\frac{1}{2}(1.5)(10)^2=75 m](https://tex.z-dn.net/?f=s%3D0%2B%5Cfrac%7B1%7D%7B2%7D%281.5%29%2810%29%5E2%3D75%20m)
(d) See attached graphs
In a uniformly accelerated motion:
- The distance travelled (x) follows the equation mentioned in part c,
![x=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=x%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
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
(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.