Question

A cheetah can accelerate from rest to a speed of 21.5 m/s in 6.75 s. What is its acceleration? m/s^2

Answer 1

`a=3.185(m)/(s^2)`

Step-by-step explanation:

Acceleration is the change in velocity for a given period of time, we can express this in the next formula:

`a = (\Delta v)/(\Delta t) =(v_(1)-v_(0))/(t_(1)-t_(0))`

In this case the values are:

`v_(0)=0\\v_(1)= 21.5 m/s\\t_(0)=0\\t_(1)= 6.75 s\\`

Inserting known values, the acceleration is:

`a= (21.5 m/s)/(6.75 s) \\a=3.185(m)/(s^2)`

Answer 2

Acceleration will be
`a=3.185m/sec^2`

Step-by-step 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
`21.5=0+a* 6.75`

`a=3.185m/sec^2`