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

Answer:

$a=3.185\frac{m}{s^2}$

Explanation:

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

$a = \frac{\Delta v}{\Delta t} =\frac{v_{1}-v_{0}}{t_{1}-t_{0}}$

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:

$a= \frac{21.5 m/s}{6.75 s} \\a=3.185\frac{m}{s^2}$

Answer 2

Answer:

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

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 $21.5=0+a\times 6.75$

$a=3.185m/sec^2$