Question

In a typical greyhound race, a dog accelerates to a speed of 20 m/s over a distance of 30 m. It then maintains this speed up to 80 m and slowed down to stop at 100 m.a) What would be a greyhound's acceleration between 0 m to 30 m, 30 m to 80 m and 80 m to 100 m?b) What would be the time taken for the entire trip?

Answer 1

For the typical greyhound race we have that:

a) The acceleration of the dog for each section is:

• Between 0 m to 30 m = 6.67 m/s²

• Between 30 m to 80 m = 0 m/s²

• Between 80 m to 100 m = -10 m/s²

b) The time taken for the entire trip is 7.5 s.

a) Let's calculate the greyhound's acceleration for each section:

• Between 0 m to 30 m

$a = \frac{v_{f}^{2} - v_{0}^{2}}{2\Delta d}$   (1)

Where:

a: is the acceleration =?

$v_{f}$: is the final velocity = 20 m/s

$v_{0}$: is the initial velocity = 0 (the dog starts from rest)

Δd: is the change in distance = 30 m - 0 m = 30 m

Then, the acceleration is:

$a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} = \frac{(20 m/s)^{2}}{2*30 m} = 6.67 m/s^{2}$  

• Between 30 m to 80 m

Since the dog maintains a velocity of 20 m/s up to 80 m, the acceleration is zero. The initial and final velocity is the same, so the acceleration in eq (1) is 0.      

• Between 80 m to 100 m

In this stage, the dog slowed down to stop at 100 m. This means that the final speed is 0.

We have:

$v_{f}$ = 0 (the dog stops)

$v_{0}$ = 20 m/s (the dog maintained this velocity up to 80 m)

Δd = 100 m - 80 m = 20 m

Hence, the acceleration is (eq 1):

$a = \frac{0 - (20 m/s)^{2}}{2*20 m} = -10 m/s^{2}$

The minus sign means that the dog is decelerating in the final section.

b) The time taken for the entire trip can be calculated as follows:

$t = t_{0-30} + t_{30-80} + t_{80-100}$  (2)

Hence, the time of each stage is:

• Between 0 m to 30 m

The time can be found with the following equation:

$t_{0-30} = \frac{v_{f} - v_{0}}{a} = \frac{20 m/s}{6.67 m/s^{2}} = 3.00 s$

• Between 30 m to 80 m

Since the acceleration is 0, the time can be calculated with the next kinematic equation:

$\Delta d = v_{0}t - \frac{1}{2}at^{2}$

$t_{30-80} = \frac{d_{f} - d_{i}}{v_{0}} = \frac{80 m - 30 m}{20 m/s} = 2.5 s$

• Between 80 m to 100 m

The time is:

$t_{80-100} = \frac{v_{f} - v_{0}}{a} = \frac{0 - 20 m/s}{-10 m/s^{2}} = 2.00 s$

Therefore, the time taken for the entire trip is (eq 2):

$t = t_{0-30} + t_{30-80} + t_{80-100} = 3.00 s + 2.5 s + 2.00 s = 7.5 s$                  

You can find another example of calculation of acceleration here: https://brainly.com/question/19867449?referrer=searchResults              

I hope it helps you!