Respuesta :
Answer:
Acceleration will be [tex]a=3.185m/sec^2[/tex]
Explanation:
We have given final velocity v = 21.5 m/sec
Time t = 6.75 sec
As cheetah starts from rest so initial velocity u = 0 m/sec
From first equation of motion we know that v = u+at, here v is final velocity, u is initial velocity, a is acceleration and t is time
So [tex]21.5=0+a\times 6.75[/tex]
[tex]a=3.185m/sec^2[/tex]
Answer:
[tex]a=3.185\frac{m}{s^2}[/tex]
Explanation:
Acceleration is the change in velocity for a given period of time, we can express this in the next formula:
[tex]a = \frac{\Delta v}{\Delta t} =\frac{v_{1}-v_{0}}{t_{1}-t_{0}}[/tex]
In this case the values are:
[tex]v_{0}=0\\v_{1}= 21.5 m/s\\t_{0}=0\\t_{1}= 6.75 s\\[/tex]
Inserting known values, the acceleration is:
[tex]a= \frac{21.5 m/s}{6.75 s} \\a=3.185\frac{m}{s^2}[/tex]
