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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 14

Calculate the average acceleration of an airplane during landing if it touches down with a velocity of 60 m/s north and comes to a complete stop over a distance of 350 m.   (2 marks)

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\(5.14\ \text{ms}^{-2}\) to the south.

Show Worked Solution
\(v^2\) \(=u^2 +2as\)  
\(a\) \(=\dfrac{v^2-u^2}{2s}\)  
  \(=\dfrac{0-60^2}{2 \times 350}\)  
  \(=-5.14\ \text{ms}^{-2}\)  

 

  • The average acceleration of the airplane is 5.14 ms\(^{-2}\) to the south.

PHYSICS, M1 EQ-Bank 10

A tennis ball machine is being tested by launching a ball straight upward from a height of 1.2 meters. The ball rises to its maximum height and then falls back down, being caught by the machine at the same height of 1.2 meters from where it was launched.

The ball had an initial velocity of 25 ms\(^{-1}\). Assume that air friction was negligible and gravity has a downward acceleration of 10 ms\(^{-2}\).

  1. Determine the total distance the ball travelled.   (2 marks)

  2. Determine the total flight time of the ball.   (2 marks)

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a.   \(62.5\ \text{m}\)

b.   \(5\ \text{sec}\)

Show Worked Solution

a.   \(\text{Define upwards as the positive direction.}\)

\(\text{At the top of the ball’s flight}\ \ \Rightarrow \ v=0\)

\(\text{Since no time of deceleration is given,}\)

\(\text{Using}\ \ v^2=u^2-2as:\)

\(0\) \(=25^{2}-2 \times -10 \times s\)  
\(s\) \(=\dfrac{25^{2}}{2 \times 10}\)  
  \(=31.25\ \text{m}\)  

 
\(\therefore\ \text{Total distance travelled}\ = 2 \times 31.25 = 62.5\ \text{m}\)
 

b.   \(\text{Find the final velocity of the ball:}\)

\(\text{Since motion is symmetrical}\ \ \Rightarrow \ \ v=-25\ \text{ms}^{-1}\)

\(\text{Find the total flight time:}\)

\(v\) \(=u + at\)  
\(-25\) \(=25-10 \times t\)  
\(t\) \(=\dfrac{25+25}{10}\)  
  \(=5\ \text{sec}\)  

PHYSICS, M1 EQ-Bank 9

An oil tanker is travelling eastward along a straight shipping route at an initial velocity of 30 ms\(^{-1}\). The captain then applies the engines in reverse, causing the tanker to slow down at a constant rate until its velocity decreases to 12 ms\(^{-1}\).

If it travelled 9000 metres while decelerating, determine

  1. the magnitude and direction of the tanker's acceleration. (2 marks)

  2. the time over which the tanker changed its velocity. (2 marks)

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a.   \(a=-0.042\ \text{ms}^{-2}\)

b.   \(674\ \text{sec}\)

Show Worked Solution

a.   \(\Delta v = 30-12=-18\ \text{ms}^{-1}\)

\(s=9\ \text{km}\ = 9000\ \text{m}\)

\(\text{Since no time of deceleration is given,}\)

\(\text{Using}\ \ v^2=u^2-2as:\)

\(a=\dfrac{v^2-u^2}{2s} = \dfrac{12^2-30^2}{2 \times 9000} = -0.042\ \text{ms}^{-2}\)

\(\therefore a=0.042\ \text{ms}^{-2}\ \text{in the opposite direction to initial velocity.}\)
 

b.   \(\text{Find time of deceleration:}\)

\(v\) \(=u + at\)  
\(t\) \(=\dfrac{v-u}{a}\)  
  \(=\dfrac{12-30}{-0.042}\)  
  \(=428.6\ \text{sec (1 d.p.)}\)  

PHYSICS, M1 EQ-Bank 8

A drone is travelling in a straight line at 56 ms\(^{-1}\) and then slows down at a constant rate until its velocity reaches 18 ms\(^{-1}\).

If it travelled 400 metres while decelerating, determine

  1. the magnitude and direction of the drone's acceleration.   (2 marks)

  2. the time over which the drone changed its velocity.   (2 marks)

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a.  \(a=-3.515\ \text{ms}^{-2}\)

b.   \(10.8\ \text{sec}\)

Show Worked Solution

a.   \(\Delta v = 56-18=-38\ \text{ms}^{-1}\)

\(\text{Since no time of deceleration is given,}\)

\(\text{Using}\ \ v^2=u^2-2as:\)

\(a=\dfrac{v^2-u^2}{2s} = \dfrac{18^2-56^2}{2 \times 400} = -3.515\ \text{ms}^{-2}\)

\(\therefore a=3.515\ \text{ms}^{-2}\ \text{in the opposite direction to initial velocity.}\)
 

b.   \(\text{Find time of deceleration:}\)

\(v\) \(=u + at\)  
\(t\) \(=\dfrac{v-u}{a}\)  
  \(=\dfrac{18-56}{-3.515}\)  
  \(=10.8\ \text{sec (1 d.p.)}\)