Question

A car starts from rest and accelerates uniformly overtime of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car in m/s^2

Answer 1

`8.10(m)/(s^2)`

Step-by-step explanation

to solve this we need to use the expression

`\begin{gathered} x=v_it+(1)/(2)at^2 \\ \text{where} \\ vf\text{ is the final sp}eed \\ vi\text{ is the initial spe}ed \\ a\text{ is the acceleration} \\ x\text{ is the traveled distance} \end{gathered}`

Step 1

Let

`\begin{gathered} vi=0(\text{ it starts from the rest)} \\ a=a \\ vt=5.21\text{ s} \\ x=110\text{ m} \end{gathered}`

replace

`\begin{gathered} x=v_it+(1)/(2)at^2 \\ 110m=(0)(2.51s)+(1)/(2)(a)(5.21s)^2 \\ 110m=0+(1)/(2)(a)(5.21s)^2 \\ 110m=13.57s^2\cdot a \\ \text{divide both sides by }13.57s^2 \\ (110m)/(13.57s^2)=(13.57s^2\cdot a)/(13.57s^2) \\ 8.10(m)/(s^2)=a \end{gathered}`

therefore, the answer is

`8.10(m)/(s^2)`

I hope this helps you