A bus starts from rest at a bus stop and accelerates at a rate of 1.12 m/s^2 for 11.0s. It runs at constant speed for 80.0 s and slows at a rate 1.20 m/s^2 until it stops at the next bus stop. (a) What is the maximum speed attained. (b) Find the total distance covered.
vmax is just after the 11seconds
vmax= a*t
distance covered:
break it up into three times
accelerating
constant speed
deaccelerating.
To solve this problem, we need to break it down into three parts: the initial acceleration, the constant speed phase, and the final deceleration.
Part 1: Initial acceleration
We are given that the bus starts from rest and accelerates at a rate of 1.12 m/s^2 for 11.0 s. To find the maximum speed attained, we can use the formula for velocity (v) as a function of time (t) during constant acceleration:
v = u + at
Where:
- v is the final velocity,
- u is the initial velocity (which is 0 m/s in this case since the bus starts from rest),
- a is the acceleration,
- t is the time.
In this case, we have u = 0, a = 1.12 m/s^2, and t = 11.0 s. Plugging these values into the equation, we get:
v = 0 + 1.12 * 11.0 = 12.32 m/s
So, the maximum speed attained during the initial acceleration phase is 12.32 m/s.
Part 2: Constant speed phase
During the constant speed phase, the velocity of the bus remains constant. To find the distance covered during this phase, we can use the formula:
distance = speed * time
In this case, the speed remains constant at 12.32 m/s for a duration of 80.0 s. Plugging these values into the equation, we get:
distance = 12.32 * 80.0 = 985.6 m
So, the distance covered during the constant speed phase is 985.6 m.
Part 3: Final deceleration
The bus slows down at a rate of 1.20 m/s^2 until it comes to a stop. Using the same velocity equation as before, but this time with v = 0, we can find the time it takes for the bus to stop:
0 = 12.32 + (-1.20) * t
Simplifying the equation, we have:
-1.20t = -12.32
Dividing both sides by -1.20, we get:
t = 10.27 s
Now, we can find the distance covered during deceleration using the distance formula:
distance = speed * time
In this case, the speed remains constant at 12.32 m/s for a duration of 10.27 s. Plugging these values into the equation, we get:
distance = 12.32 * 10.27 = 126.53 m
So, the distance covered during the deceleration phase is 126.53 m.
Total distance covered:
To find the total distance covered by the bus, we sum up the distances covered during each phase:
Total distance = distance of initial acceleration + distance of constant speed phase + distance of deceleration
= 0.5 * 1.12 * (11.0)^2 + 985.6 + 0.5 * 1.20 * (10.27)^2
= 67.84 + 985.6 + 62.73
= 1116.17 m
Therefore, the total distance covered by the bus is 1116.17 m.