Question

Please need help fast

Answer 1

(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`

The average acceleration of the horse can be calculated as:

`a=(v-u)/(t)`

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=(15-0)/(10)=1.5 m/s^2`

(c) 75 m

For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:

`s=ut+(1)/(2)at^2`

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`

Substituting,

`s=0+(1)/(2)(1.5)(10)^2=75 m`

(d) See attached graphs

In a uniformly accelerated motion:

- The distance travelled (x) follows the equation mentioned in part c,

`x=ut+(1)/(2)at^2`

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

`a=1.5 m/s^2` (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.