A bus starts from rest and accelerates uniformly for 20 seconds to a speed of 18 km/h. It then moves with a uniform velocity and it is finally brought to rest in 100 meteres with a constant retardation. If the total distance travelled is 0.5km. Find the accelration, dation and total time taken. My answers are total time = 110s Retardation -1/4ms^-2. Please check if they r correct or not!!!
Not correct. I will be happy to critique your thinking.
then how to do it!!!
an aeroplane takes of at an angle of 45deg to the horizontal. If the vertical component of its velocity is 300kmph, Calculate its actual velocity. what is the horizontal component of it velocity?
i studied that in 6th grade lol
To find the acceleration, retardation, and total time taken, let's break down the problem into three segments:
1. Acceleration:
The bus starts from rest and accelerates uniformly to a speed of 18 km/h in 20 seconds. To find the acceleration, we'll convert the speed from km/h to m/s:
18 km/h = (18 * 1000) / 3600 m/s = 5 m/s
Using the equation v = u + at, where v is the final velocity, u is the initial velocity (0 m/s since the bus starts from rest), a is the acceleration, and t is the time, we can rearrange the equation to find the acceleration:
5 m/s = 0 m/s + a * 20 s
Simplifying the equation:
a = 5 m/s / 20 s
a = 0.25 m/s²
So, the acceleration of the bus is 0.25 m/s².
2. Uniform Velocity:
The bus moves with a uniform velocity after reaching 18 km/h. We need to determine the distance covered during this period. The total distance traveled is given as 0.5 km, which corresponds to 500 meters.
To find the distance covered while moving with uniform velocity, we subtract the distances covered during acceleration and deceleration from the total distance:
Distance while moving with uniform velocity = Total distance - Distance covered during acceleration - Distance covered during deceleration
Distance while moving with uniform velocity = 500 m - (0.5 * 100) m - (0.1 * 100) m (converting 0.5 km and 100 meters to meters)
Distance while moving with uniform velocity = 500 m - 50 m - 10 m
Distance while moving with uniform velocity = 440 m
So, the distance covered while moving with uniform velocity is 440 meters.
3. Deceleration:
The bus finally comes to rest in 100 meters with a constant retardation (deceleration). We can use the equation v² = u² + 2as, where v is the final velocity (0 m/s at rest), u is the initial velocity, a is the retardation, and s is the distance.
Rearranging the equation to find the retardation:
0 m/s = u² + 2 * a * 100 m
Simplifying the equation:
- 2 * a * 100 m = u²
Since we know that the initial velocity u = 18 km/h and need to convert it to m/s:
u = (18 * 1000) / 3600 m/s = 5 m/s
Substituting the values into the equation:
-2 * a * 100 m = (5 m/s)²
-200a = 25 m²/s²
a = -25 m²/s² / -200
a = 0.125 m/s²
So, the retardation (deceleration) of the bus is 0.125 m/s².
Total Time Taken:
To find the total time taken, we sum up the individual times for acceleration, uniform velocity, and deceleration:
Total Time Taken = Time for acceleration + Time for uniform velocity + Time for deceleration
Total Time Taken = 20 s + (440 m / 5 m/s) + (100 m / (5 m/s + (-0.125 m/s²) * t))
Calculating the total time taken:
Total Time Taken = 20 s + 88 s + (100 m / 4.875 m/s)
Total Time Taken = 20 s + 88 s + 20.51 s
Total Time Taken ≈ 128.51 s
Therefore, the correct answers are:
Acceleration = 0.25 m/s²
Retardation = -0.125 m/s²
Total Time Taken ≈ 128.51 s