Question

Can someone pls explain this question to me. Thanks in advance! ​

Answer 1

Answer: $7.407 m/s^{2}$

Explanation:

In this question we are asked to find the cheeta's accelaration and we assume it is constant. Therefore, we can use the folowing equation:

$V=V_{o}+at$ (1)

Where:

$V=80 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=22.22 m/s$ is the cheeta's final speed

$V_{o}= 0 m/s$ is the cheeta's initial speed

$t=3 s$ is the time in which the cheeta passes from $V_{o}$ to $V$

$a$ is the cheeta's acceleration

Isolating $a$ from (1):

$a=\frac{V}{t}$ (2)

$a=\frac{22.22 m/s}{3 s}$ (3)

Finally:

$a+7.407 m/s^{2}$