Cheetahs can accelerate to a speed of 19.3 m/s19.3 m/s in 2.55 s2.55 s and can continue to accelerate to reach a top speed of 31.0 m/s31.0 m/s . assume the acceleration is constant until the top speed is reached and is zero thereafter. let the +x+x direction point in the direction the cheetah runs. express the cheetah's top speed in miles per hour (mi/h)(mi/h) . =vtop= mi/hmi/h starting from a crouched position, how much time does it take a cheetah to reach its top speed and what distance does it travel in that time? =taccel= ss =d= mm if a cheetah sees a rabbit 137.0 m137.0 m away, how much time will it take to reach the rabbit, assuming the rabbit does not move and the cheetah starts from rest?

TopSpeed (in mph=mi/h)=69,3452 mi/h   T_top speed=4,0958 seconds and when reaching top speed the distance traveled is 63,4838 m. Time to 137m is 6.4672 s

Explanation:

We can use a rule of three to turn one unit of measurement into other, now we know that

1 mile=1609,34 m

and

1 hour = 3600 seconds,

then we have that

Also we have enough information to obtain the constant acceleration rate since, vf=vo+a*t, then a=(vf-vo)/t, we are told that the Cheetahs accelerate from 0 to 19.3 m/s in 2.55 s, then

a=(19,3m/s-0m/s)/2,55 s=7,5686 m/s^2

With this information we can also get the time to topspeed, since vf=vo+a*t again, we have now for the time t:

t=(vf-vo)/a

, which is the time to topspeed. The distance traveled is also found using the formula d=xo+1/2*a*t^2, then

, which is the distance at which the cheetahs reach the topspeed.

Finally to get the time it needs to travel 137 m, we remember what it says that after reaching the top speed it does not longer accelerate, then the rest of the

73,5161 m(=137m-63,4838m) are traveled at 31m/s, then we use the formula for constant speed for those 73,5161 m, then we know x=v*t, then t=x/v

here

Then the total time to travel the 137 m is 2,3714 s at constant topspeed of 31m/s+ 4,0958 s that it takes for the cheetah to attain the 31m/s, that is, 6.4672s