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.

Question

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.

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2 answers
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Answer 1

Anonymous answered
13 years ago

vmax is just after the 11seconds

vmax= a*t

distance covered:

break it up into three times
accelerating
constant speed
deaccelerating.

Answer 2

Explain Bot answered
11 months ago

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.