❤️
Q1.a scooter moving at the spped of 12 metre per second is stopped by applying breaks which produces uniform retardation of 0.6 metre per second square.how much distance will be covered by the scooter till it stops
Answers
Answered by
31
Answer:
Using the standard equations of motion we know:
Dist(x)= Initial Velocity x Time + 1/2 Constant Acceleration x Time^2 (Squared)
We know the Initial Velocity (12 .0 m/s) and the Constant Acceleration (-0.6 m/s/s negative for deceleration) but not the time.
But we also know:
Change in Velocity (∆V) = Acceleration x Time (at)
So
∆V = 12 m/s - 0 m/s = a x t = 0.6 m/s^2 x t
Solving for t = (12 m/s)/(0.6 m/s/s) = 20 s
Plugging that into equation 1:
Dist = 12 m/s x 20 s + 1/2 (-0.6m/s/s) x (20 s)^ 2 = 240 m - 120 m = 120 m to stop
Explanation:
Hiii!!!
Prince...
Good night.
¯\_(ツ)_/¯¯\_(ツ)_/¯
Be my friend please....
Answered by
3
Answer:
tumhe yaad hai na.. 6 march ko mera birthday hai... aaoge na tb??.. plz
Similar questions
Computer Science,
11 days ago
English,
11 days ago
English,
11 days ago
Social Sciences,
23 days ago
English,
23 days ago
Chemistry,
9 months ago
Physics,
9 months ago