Question

Pls answer fast. I will mark ur ans as brainiest....
A particle moves from a to b with speed of 120km/hr and returns from b to a with speed of 80km/hr find average speed and velocity

Answer 1

Distance between the two points is the same in case of departing and arriving.

V1=120 KMPH

V2=80 KMPH

In case of equal distances, the average speed is

$\frac{2v1.v2}{v1 + v2}$

=

$\frac{2 \times 120 \times 80}{120 + 80}$

=96 KMPH.

Average. velocity =0

Answer 2

Let us assume that the distance travelled by a particle is 'x'.

Given that, a particle moves from a to b with speed of 120km/hr and returns from b to a with speed of 80km/hr. (s1 = 120 km/hr and s2 = 80 km/hr)

We have to find the average speed and average velocity of the particle.

Average speed is defined as the total distance covered by a particle with respect to total time taken.

Whereas average velocity is as the total displacement with respect to total time taken.

For average speed:

Average speed = (2 × s1 × s2)/(s1 + s2)

= (2 × 120 × 80)/(120 + 80)

= 19200/200

= 96

Therefore, the average speed of the particle is 96 km/hr.

For average velocity:

Since particle is starting from a and moves to b & then return backs to a. As it's initial and final points are same. So, the displacement of the particle is zero.

From a to b:

t1 = d/120

From b to a:

t2 = d/80

Average velocity = Displacement/Time

= 0/(d/120 + d/80)

= 0

Therefore, the average velocity of the particle is 0 km/hr.