Question

a cyclist took 80 s to slow down from 12m/s to 0 m/s what is distance travelled

Answer 1

Provided that:

• Time = 80 seconds
• Final velocity = 0 mps
• Initial velocity = 12 mps

To calculate:

• Distance travelled

Solution:

• Distance = 480 m

Using concepts:

• To solve this question firstly there is a need to find out the acceleration. We can find acceleration by using either first equation of motion or acceleration formula.

• Choice may vary!

• We can use either second equation of motion or third equation of motion to calculate the distance travelled.

• Choice may vary!

Using formulas:

• Three equations of motion are mentioned below respectively:

$\begin{gathered}\boxed{\begin{array}{c}\\ {\pmb{\sf{Three \: equations \: of \: motion}}} \\ \\ \sf \star \: v \: = u \: + at \\ \\ \sf \star \: s \: = ut + \: \dfrac{1}{2} \: at^2 \\ \\ \sf \star \: v^2 - u^2 \: = 2as\end{array}}\end{gathered}$

• Acceleration formula:

• ${\small{\underline{\boxed{\pmb{\sf{a \: = \dfrac{v-u}{t}}}}}}}$

Required solution:

~ Firstly let us calculate the acceleration!

By using acceleration formula...

$:\implies \sf a \: = \dfrac{dv}{dt} \\ \\ :\implies \sf a \: = \dfrac{\Delta \: v}{t} \\ \\ :\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf Acceleration \: = \dfrac{v-u}{t} \\ \\ :\implies \sf Acceleration \: = \dfrac{0-12}{80} \\ \\ :\implies \sf Acceleration \: = \dfrac{-12}{80} \\ \\ :\implies \sf Acceleration \: = \dfrac{-6}{40} \\ \\ :\implies \sf Acceleration \: = \dfrac{-3}{20} \\ \\ :\implies \sf Acceleration \: = -0.15 \: ms^{-2}$

Negative sign indicates that the speed of the cyclist slows down.

By using first equation of motion...

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 0 = 12 + a(80) \\ \\ :\implies \sf 0 - 12 = 80a \\ \\ :\implies \sf -12 = 80a \\ \\ :\implies \sf Acceleration \: = \dfrac{-12}{80} \\ \\ :\implies \sf Acceleration \: = \dfrac{-6}{40} \\ \\ :\implies \sf Acceleration \: = \dfrac{-3}{20} \\ \\ :\implies \sf Acceleration \: = -0.15 \: ms^{-2}$

~ Now let us find out the distance!

By using third equation of motion...

$:\implies \sf v^2 - u^2 \: = 2as \\ \\ :\implies \sf (0)^{2} - (12)^{2} = 2(-0.15)(s) \\ \\ :\implies \sf 0 - 144 = -0.30s \\ \\ :\implies \sf -144 = -0.30s \\ \\ :\implies \sf 144 = 0.30s \\ \\ :\implies \sf \dfrac{144}{0.30} \: = s \\ \\ :\implies \sf \dfrac{14400}{30} \: = s \\ \\ :\implies \sf \dfrac{1440}{3} \: = s \\ \\ :\implies \sf 480 \: = s \\ \\ :\implies \sf s \: = 480 \: m \\ \\ :\implies \sf Distance \: = 480 \: m$

By using second equation of motion...

$:\implies \sf s \: = ut + \: \dfrac{1}{2} \: at^2 \\ \\ :\implies \sf s \: = 12(80) + \dfrac{1}{2} \times -0.15(80)^{2} \\ \\ :\implies \sf s \: = 960 + \dfrac{1}{2} \times -0.15(6400) \\ \\ :\implies \sf s \: = 960 + \dfrac{1}{2} \times -960 \\ \\ :\implies \sf s \: = 960 + 1 \times -480 \\ \\ :\implies \sf s \: = 960 + (-480) \\ \\ :\implies \sf s \: = 960 - 480 \\ \\ :\implies \sf 480 \: = s \\ \\ :\implies \sf s \: = 480 \: m \\ \\ :\implies \sf Distance \: = 480 \: m$