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Artemon [7]
2 years ago
10

A passenger bus is travelling 28.0 m/s to the right when the driver applies the brakes. The bus stops in 5.00 s. What is the acc

eleration of the bus as it comes to a stop?
Physics
2 answers:
zaharov [31]2 years ago
6 0

Answer:  - 5.60 m/s^2

Explanation:

Speed of passenger bus = 28.0 m/s to the right  

Time took for the bus stops = 5.00 s.

What is the acceleration of the bus as it comes to a stop = x

Step 1:

Calculate the change in velocity  

d = v2 (Final velocity)  - v1 (Initial Velocity)

d= 0 – 28

d= -28m/s  

Acceleration=  Change in velocity / time taken

Acceleration = -28 m/s / 5

Acceleration = - 5.60 m/s^2 to the right

Note: Do not forget the units, otherwise they make cost you marks!


MAVERICK [17]2 years ago
3 0
Change in velocity = d(v)
d(v) = v2 - v1 where v1 = initial speed, v2 = final speed
v1 = 28.0 m/s to the right
v2 = 0.00 m/s
d(v) = (0 - 28)m/s = -28 m/s to the right

Change in time = d(t)
d(t) = t2 - t1 where t1 = initial elapsed time, t2 = final elapsed time
t1 = 0.00 s
t2 = 5.00 s
d(t) = (5.00 - 0.00)s = 5.00s

Average acceleration = d(v) / d(t)
(-28.0 m/s) / (5.00 s)
(-28.0 m)/s * 1 / (5.00 s) = -5.60 m/s² to the right
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<span>(a) -9.97 m/s
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   This is a simple problem in integral calculus. You've been given part of the 2nd derivative (acceleration), but not quite. You've been given the force instead. So let's setup a function for acceleration.

   f''(x) = -8x N / 3.1 kg= -8x kg*m/s^2 / 3.1 kg = -2.580645161x m/s^2

   So the acceleration of the body is now expressed as
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    Let's calculate the anti-derivative from that.
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   Now let's use the known velocity value at x = 2.0 to calculate C
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   So the velocity is -9.97 m/s

   (b) we want a velocity of 5.8 m/s
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 0 = -1.290322581x^2 + 10.36129032
 1.290322581x^2 = 10.36129032
 x^2 = 8.029999998
 x = 2.833725463</span>
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