Two cars x and y travelling side by side are subjected to constant retardations 1/2m/s and 1/6m/s.After passing a point A,they come to rest 300m before and 150m beyond a second point B.Find common velocity at A and distance AB
This term "retardations"....
It has units of velocity. Velocity is not a "retardation", see Newtons first law of motion.
I do not know what you are asking.
sorry, I mean m/s^2
To find the common velocity at point A and distance AB, we need to analyze the motion of the two cars.
Let's consider car X and car Y individually.
For Car X:
- Retardation (a) = -1/2 m/s² (negative since it's a deceleration)
- Initial velocity (u₁) = unknown
- Final velocity (v₁) = 0 m/s (since it comes to rest)
- Distance between A and B (s₁) = 300 m
Using the first equation of motion: v₁² = u₁² + 2as₁,
we can substitute the known values to solve for u₁:
0 = u₁² + 2*(-1/2)*300
0 = u₁² - 300
u₁² = 300
u₁ = √300 ≈ 17.32 m/s
Similarly, for Car Y:
- Retardation (a) = -1/6 m/s²
- Initial velocity (u₂) = unknown
- Final velocity (v₂) = 0 m/s (since it comes to rest)
- Distance between A and B (s₂) = 150 m
Using the same equation of motion: v₂² = u₂² + 2as₂,
we can substitute the known values to solve for u₂:
0 = u₂² + 2*(-1/6)*150
0 = u₂² - 50
u₂² = 50
u₂ = √50 ≈ 7.07 m/s
Now, since both cars start at point A, their initial velocities are equal. Therefore, the common velocity at point A is:
Common velocity at A = u₁ = u₂ ≈ 17.32 m/s
The distance AB is the sum of the distances traveled by both cars:
AB = 300m + 150m = 450m
So, the common velocity at point A is approximately 17.32 m/s and the distance AB is 450m.