Question

A truck covers 40.0 m in 7.10 s while uniformly slowing down to a final velocity of 1.25 m/s.(a) Find the truck's original speed. m/s(b) Find its acceleration. m/s2

Answer 1

original speed: 10 m/s

acceleration : -1.23 m(s^2) , or desaccelerating ar a t rate of 1.23 m/s^2

Explanation

to solve this we need the formuilas

$\begin{gathered} v_f=v_0-at \\ \\ v_{f^2}=v^2_0-2ad \end{gathered}$

Step 1

Let

$\begin{gathered} \text{distance =d= 40 m} \\ \text{time}=\text{ t=7.1 s} \\ \text{ final velocity = v}_f=1.25\text{ }\frac{m}{s} \\ \text{acceleration}=a \\ v_o \end{gathered}$

now, replace and solve for a

$\begin{gathered} v_f=v_0-at \\ 1.25\text{ }\frac{m}{s}=v_0-a(7.1\text{ s)} \\ 1.25=v_0-7.1a\rightarrow equation(1) \end{gathered}$

and,

$\begin{gathered} v^2_f=v^2_0-2ad \\ \text{replace} \\ 1.25^2=v^2_0-80a \\ 1.25^2=v^2_0-80a\rightarrow equation(2) \\ \end{gathered}$

Step 2

solve the equations

$\begin{gathered} 1.25=v_0-7.1a\rightarrow equation(1) \\ 1.25^2=v^2_0-80a\rightarrow equation(2) \end{gathered}$

a) isolate the a value in equaion (1) and then replace in equation (2)

$\begin{gathered} 1.25=v_0-7.1a\rightarrow equation(1) \\ 1.25-v_0=-7.1a \\ \frac{1.25-v_0}{-7.1}=a \end{gathered}$

now,set equal in equation (2)

$\begin{gathered} 1.5625=v^2_0-80a\rightarrow equation(2) \\ 1.5625=v^2_0-80(\frac{1.25-v_0}{-7.1}) \\ 1.5625=v^2_0+\frac{100-80v_0}{-7.1} \\ 1.5625=v^2_0+\frac{100}{7.1}-\frac{80v_0}{7.1} \\ 1.5625=v^2_0+14.08-11.26v_0 \\ 0=v^2_0+14.08+11.26v_0-1.5625 \\ 0=v^2_0+11.26v_0-12.5175 \\ v_0=10\text{ }\frac{m}{s} \end{gathered}$

b) finally, replace the v0 value to find the acceleration

$\begin{gathered} \frac{1.25-v_0}{-7.1}=a \\ \text{replace} \\ \frac{1.25-10}{-7.1}=a \\ a=\frac{-8.75}{-7.1}=1.23\text{ }\frac{m}{s^2} \end{gathered}$

therefore,

original speed: 10 m/s

acceleration : -1.23 m(s^2) , or desaccelerating ar a t rate of 1.23 m/s^2