The maximum speed of a bus is 72km/h, it accelerates uniformly at the rate of 1m/s^2 and s uniformly at the rate of 4m/s^2. Find the least time in which it can do a journey of 1km ?

break the problem into three parts:

accelerating, constant speed, deaccelerating.

Let x be the accelerating distance, y be the constant speed distance, x/4 be the deaccelerating distance.


1000=x+y+x/4 or 5x/4 +y=1000, or y= 1000-5x/4

so solve for x first. 72km/hr = 20m/s
Vf^2=2ad
400=2*1*x or x=200 that leads to all the other distances.

time= sum of distances divided by avg velocity during that time.

time=200/10 +750/20 +50/10

check my work.

To find the least time in which the bus can complete a journey of 1 km, we can divide the journey into two parts: acceleration and retardation.

First, let's calculate the time taken during the acceleration phase. We know that acceleration is given as 1 m/s^2. We also know that the final velocity during this phase is 72 km/h. Since we need to work with consistent units, we convert the final velocity to m/s:

72 km/h * (1000 m/1 km) * (1 h/3600 s) = 20 m/s

Now, we can use the equation of motion to find the time taken during acceleration:

v = u + at

Here, v is the final velocity (20 m/s), u is the initial velocity (0 m/s), a is the acceleration (1 m/s^2), and t is the time taken during acceleration (which is what we need to find).

Substituting the values into the equation, we get:

20 m/s = 0 m/s + 1 m/s^2 * t

Simplifying the equation:

20 m/s = t

Therefore, the time taken during acceleration is 20 seconds.

Next, let's calculate the time taken during the retardation phase. We know that the retardation (deceleration) is 4 m/s^2. The final velocity during this phase is 0 m/s, as the bus comes to a stop.

Using the same equation of motion, but this time with a negative acceleration:

v = u + at

Here, v is the final velocity (0 m/s), u is the initial velocity (20 m/s), a is the acceleration (-4 m/s^2), and t is the time taken during retardation (which is what we need to find).

Substituting the values into the equation, we get:

0 m/s = 20 m/s + (-4 m/s^2) * t

Simplifying the equation:

0 m/s = 20 m/s - 4 m/s^2 * t
4 m/s^2 * t = 20 m/s
t = 20 m/s / 4 m/s^2
t = 5 seconds

Therefore, the time taken during retardation is 5 seconds.

Finally, we add the time taken during acceleration and retardation to get the total time:

Total time = Time taken during acceleration + Time taken during retardation
Total time = 20 seconds + 5 seconds
Total time = 25 seconds

Hence, the least time in which the bus can complete a journey of 1 km is 25 seconds.

The answer to this question is 62.5s. Can you please explain it properly what is that Vf^2 = 2ad