Question

A car travels from stop A to stop B with a speed of 30 km/h and then returns back to A with a speed of 50 km/h. Find (i) displacement of the car (ii) average speed of the car.

Answer 1

Answer:

Displacement= 0

Average speed = 37.5 km/h

Explanation:

As per the provided information in the given question, we have:

• A car travels from stop A to stop B with a speed of 30 km/h.
• Then returns back to A with a speed of 50 km/h.

We are asked to calculate the displacement of the car and average speed of car.

★ Calculating displacement :

We know that, whenever the body comes to its initial position after covering a certain distance or when its initial point and final point is same ; then displacement is 0.

Here, A is the car's initial position and B is its final position. Since, it came back to its initial point (A) after covering certain distance, so the displacement is 0.

$\\ \longrightarrow \bf{\quad \underline{ Displacement = 0 \; km}}$

★ Calculating average speed :

$\\ \longrightarrow \quad \pmb{\boxed{\sf {Speed_{(avg)} = \dfrac{Total \; Distance}{Total \; time} }} }$

Finding total distance :

Let us assume the distance from A to B as d km. So,

$\\ \longrightarrow \sf{\quad {Total \; distance = AB + BA }}$

$\\ \longrightarrow \sf{\quad {Total \; distance = (d+ d) \; km }}$

$\\ \longrightarrow \sf{\quad {Total \; distance = 2d \; km }}$

Finding total time :

Let time taken to cover the distance from A to B be $\sf (t_1)$ and time taken to cover the distance from B to A be $\sf (t_2)$ .

$\\ \longrightarrow \sf{\quad { t_1 = \dfrac{Distance_{(A \; to \; B)}}{Speed_{(A \; to \; B)}} }}$

$\\ \longrightarrow \sf{\quad {t_1 = \dfrac{d}{30} hrs}}$

Also,

$\\ \longrightarrow \sf{\quad { t_2 = \dfrac{Distance_{(B \; to \; A)}}{Speed_{(B \; to \; A)}} }}$

$\\ \longrightarrow \sf{\quad {t_2 = \dfrac{d}{50} hrs}}$

Therefore,

$\\ \longrightarrow \sf{\quad { Total \; time = t_1 + t_2}}$

$\\ \longrightarrow \sf{\quad { Total \; time = \Bigg ( \dfrac{d}{30} +\dfrac{d}{50} \Bigg ) \; hrs}}$

$\\ \longrightarrow \sf{\quad { Total \; time = \Bigg ( \dfrac{5d+3d}{150} \Bigg ) \; hrs}}$

$\\ \longrightarrow \sf{\quad { Total \; time = \dfrac{8d}{150} \; hrs}}$

Now, substitute the values of total time and total distance in the formula of average speed.

$\\ \longrightarrow \quad \pmb{\boxed{\sf {Speed_{(avg)} = \dfrac{Total \; Distance}{Total \; time} }} }$

$\\ \longrightarrow \sf{\quad {Speed_{(avg)} = \dfrac{2d}{ \cfrac{8d}{150} } \; km/h }}$

$\\ \longrightarrow \sf{\quad {Speed_{(avg)} = 2d \times \dfrac{150}{8d} \; km/h }}$

$\\ \longrightarrow \sf{\quad {Speed_{(avg)} = \dfrac{150}{4} \; km/h }}$

$\\ \longrightarrow \bf{\quad \underline{Speed_{(avg)} = 37.5 \; km/h }}$

Therefore, average speed of the car is 37.5 km/h.