Cheetahs are the fastest land mammal. They can accelerate from 0 to 60
mph in 3 seconds. Assuming a cheetah’s acceleration is constant, how far
has it run during that time interval?
acceleration --- a
then
v = at + c
at the start, t = 0, and v=0
0 = 0 + c ---> c = 0
so v = at
when t = 3 sec = 3/3600 hrs or 1/1200, v = 60
60 = a(1/1200)
a = 72000 ------> v = 72000t
if s is the distance travelled ,
s = 36000t^2 + k
when t = 0 , s = 0, so k = 0
s = 36000t^2
when t = 3 sec or 1/1200 hrs,
s = 36000(1/1200)^2 = .025 miles
physics might have some basic formulas to do this, I did it by Calculus.
While traveling on the highway, a driver slows from 24 m / s to 15 m / s in 12 seconds. What is driver acceleration? (Remember that a negative value indicates a slowdown or "slowdown").
To find the distance the cheetah has run during the 3-second time interval, we need to use the formula for distance traveled with constant acceleration:
Distance (d) = Initial Velocity (v0) * Time (t) + (1/2) * Acceleration (a) * Time^2
In this case:
- The initial velocity (v0) is 0 mph, as the cheetah starts from rest.
- The time (t) is 3 seconds.
- The acceleration (a) can be found by converting the acceleration from mph to m/s^2.
First, let's convert the acceleration from mph to m/s^2:
Acceleration (a) = (Change in Velocity) / Time
= (60 mph - 0 mph) / (3 seconds)
= 20 mph/s
Now, we need to convert mph/s to m/s^2.
1 mph = 0.44704 m/s
So,
Acceleration (a) = 20 mph/s * 0.44704 m/s
= 8.9408 m/s^2
Now, substitute the values into the distance formula:
Distance (d) = (0 mph) * (3 seconds) + (1/2) * (8.9408 m/s^2) * (3 seconds)^2
Simplifying,
Distance (d) = 0 + 4.4704 m/s^2 * 9 seconds^2
= 4.4704 m/s^2 * 81 seconds^2
Calculating,
Distance (d) = 362.4192 meters
Therefore, the cheetah has run approximately 362.42 meters during the 3-second time interval.