Question

1. A car travels from a point A to point B at constant speed of 50 km/h. On return
journey along the same path, it travels at a constant speed of 40 km/h. Find its
distance travelled, displacement and average speed.​

Answer 1

★$\color{darkblue}\underline{\underline{\sf Given-}}$

• Car travels from point A to point B with speed ${\sf (v_1)}$ = 50km/h
• Return back to point A with speed ${\sf (v_2)=40km/h}$

$\color{darkblue}\underline{\underline{\sf To \: Find-}}$

• Distance
• Displacement
• Average Speed

━━━━━━━━━━━━━━━━━━━━━━━

★$\color{darkblue}\underline{\underline{\sf Solution-}}$

★ Convert km/s into m/s

$\implies{\sf v_1=50\:km/h }$

$\implies{\sf v_1=50×\dfrac{5}{18}}$

$\color{orange}\implies{\sf v_1=13.8\:m/s}$

$\implies{\sf v_2=40\:km/h}$

$\implies{\sf v_2=40×\dfrac{5}{18} }$

$\color{orange}\implies{\sf v_2=11.1\:m/s}$

❏ Distance

Let distance Travelled from point A to point B is s

therefore , Distance Travelled from point A to point B and then from point B to point A will be 2s

❏ Displacement

We know that displacement is Shortest distance Travelled by the object/person.

here , car come back to its starting position therefore , displacement will be zero.

❏ Average Speed

$\color{violet}\bullet\underline{\boxed{\sf V_{av}=\dfrac{Total\: Distance}{Total \: Time}}}$

$\implies{\sf v_{av}=\dfrac{2s}{\dfrac{s}{13}+\dfrac{s}{11.1}} }$

$\implies{\sf v_{av}=\dfrac{2s}{\dfrac{11.1s+13.8s}{153.1}} }$

$\implies{\sf v_{av}=\dfrac{2s×153.1}{24.9s}}$

$\color{red}\implies{\sf v_{av}=12.29\:m/s}$

$\color{darkblue}\underline{\underline{\sf Answer-}}$

Distance 2s

Displacement zero

Average speed 12.29 m/s