A vehicle is uniformly retarded and brought to rest from the velocity of 54km /h in 5 second find the retardation and the distance covered during this period
To find the retardation and the distance covered during this period, we can use the following formulas:
1. Retardation (a) = Change in velocity (Δv) / Time taken (Δt)
2. Distance covered (s) = Initial velocity (u) * Time taken (Δt) + 0.5 * Retardation (a) * (Time taken)^2
Given:
Initial velocity (u) = 54 km/h = 15 m/s (convert from km/h to m/s)
Time taken (Δt) = 5 seconds
1. Convert the initial velocity from km/h to m/s:
Initial velocity (u) = 54 km/h * (1 m/s / 3.6 km/h) = 15 m/s
Now, substitute the values into the first equation to find the retardation:
Retardation (a) = Change in velocity (Δv) / Time taken (Δt)
Δv = 0 m/s (since the vehicle is brought to rest)
Δt = 5 seconds
a = 0 m/s - 15 m/s / 5 seconds = -3 m/s²
The retardation is -3 m/s².
Now, substitute the values into the second equation to find the distance covered:
Distance covered (s) = Initial velocity (u) * Time taken (Δt) + 0.5 * Retardation (a) * (Time taken)^2
s = 15 m/s * 5 seconds + 0.5 * (-3 m/s²) * (5 seconds)^2
s = 75 m - 0.5 * 3 m/s² * 25 s²
s = 75 m - 37.5 m
s = 37.5 m
The distance covered during this period is 37.5 meters.