A particle moves on a straight
road covers half of the total
distance with vi. The other half
of the distance is covered in
aequal time intervals with
Speeds Ds and V3. The averge speed
of the besticle during this' motion
is.
Answers
Answer:
see the attachment ☺️✨
Explanation:
select as brianliast answer
Explanation:
The average speed of particle during this motion is 4 m/s.
Given that,
First half of a distance is covered with speed = 3 m/s
The other half of a distance is covered in 2 equal time intervals at speed of 4.5 M second and 7.5 seconds respectively
Let the total distance covered be 2 x.
Let t be the time taken the first half of the distance.
t = \dfrac{x}{3}t=
3
x
Let the other half of a distance be covered in two equal intervals of time.
x=4.5t'+7.5t'x=4.5t
′
+7.5t
′
x=12t'x=12t
′
t'=\dfrac{x}{12}t
′
=
12
x
The total time is
t''=t+2t't
′′
=t+2t
′
t''=\dfrac{x}{3}+2\times\dfrac{x}{12}t
′′
=
3
x
+2×
12
x
t''=\dfrac{x}{3}+\dfrac{x}{6}t
′′
=
3
x
+
6
x
The average speed is equal to the total distance divided by the total time.
v_{avg}=\dfrac{2x}{\dfrac{x}{3}+\dfrac{x}{6}}v
avg
=
3
x
+
6
x
2x
v_{avg}=4 m/sv
avg
=4m/s
Hence, The average speed of particle during this motion is 4 m/s.