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PHYSICS, M1 EQ-Bank 6 MC

A drag racer must accelerate from rest to a speed of 80 m/s. If the car accelerates uniformly at 4.0 m/s², what is the minimum track length required to reach this speed?

  1. 750 m
  2. 800 m
  3. 850 m
  4. 900 m
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\(B\)

Show Worked Solution
\(v^2\) \(=u^2 + 2as\)  
\(s\) \(=\dfrac{v^2-u^2}{2a}\)  
  \(=\dfrac{80^2-0^2}{2 \times 4}\)  
  \(=800\ \text{m}\)  

 

\(\Rightarrow B\)

PHYSICS, M1 EQ-Bank 5 MC

A stone is tossed vertically upward from the edge of a canyon. It rises 9 meters above the rim before falling down to the canyon floor, which is 27 meters below the rim.

What is the total distance travelled by the stone?

  1. 36 m
  2. 42 m
  3. 45 m
  4. 48 m
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\(C\)

Show Worked Solution
  • The distance the stone travels from the edge of the canyon up into the air and back down to the edge of the canyon is 9 m + 9 m = 18 m.
  • The distance travelled by the stone from the edge of the canyon to the bottom of the canyon is 27 m.
  • The total distance travelled by the stone is 18 m + 27 m = 45 m.

\(\Rightarrow C\)

PHYSICS, M1 EQ-Bank 11

A bus departs from its depot, starting from rest and accelerating uniformly at 2.0 ms\(^{-2}\) for 10 seconds until it reaches a speed of 20 ms\(^{-1}\). It then travels at this constant speed for 50 seconds before decelerating uniformly at – 2.5 ms\(^{-2}\) until coming to a complete stop at the next bus stop.

  1. In the space provided, sketch a velocity–time graph of the bus's journey, clearly marking the acceleration phase, the constant speed phase, and the deceleration phase with all key values.   (3 marks)
  1. Calculate the total distance the bus covers during this trip.   (2 marks)

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a.    

       

b.    \(1180\ \text{m}\)

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a.   
           

b.    Total distance travelled by the bus is equal to the area under the velocity-time graph.

\(\text{Distance}\) \(=(\dfrac{1}{2} \times 10 \times 20) + (50 \times 20) + (\dfrac{1}{2} \times 8 \times 20) \)  
  \(=1180\ \text{m} \)  

PHYSICS, M1 EQ-Bank 2-3 MC

Using the velocity-time graph below
 

 
Part 1

Determine the magnitude of the displacement:

  1. \(32\ \text{m}\)
  2. \(40\ \text{m}\)
  3. \(48\ \text{m}\)
  4. \(64\ \text{m}\)

 
Part 2

Determine the average acceleration between 4 and 8 seconds:

  1. \(-4\ \text{ms}^{-2}\)
  2. \(-2\ \text{ms}^{-2}\)
  3. \(-1\ \text{ms}^{-2}\)
  4. \(2\ \text{ms}^{-2}\)
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Part 1: \(C\)

Part 2: \(B\)

Show Worked Solution

Part 1

  • The displacement for the motion can be calculated by finding the area under the velocity time graph. 
  • By splitting the graph up into the square and triangle, the area under the curve is:

\(\text{Area}\ =(4 \times 8) + (\dfrac{1}{2} \times 4 \times 8) = 32 +16 = 48\ \text{m}\)

\(\Rightarrow C\)
 

Part 2

  • Average acceleration between t=4 and t=8 is:

\(a= \dfrac{\Delta v}{\Delta t} = \dfrac{0-8}{8-4} = -2\ \text{ms}^{-2}\)

\(\Rightarrow B\)