Suggested time to complete: 30 minutes

1. A car is moving at $\scriptsize 30\text{ m}\text{.}{{\text{s}}^{{-1}}}$ east when a driver steps on the brakes. The car’s velocity decreases uniformly to $\scriptsize 20\text{ m}\text{.}{{\text{s}}^{{-1}}}$ in $\scriptsize 10\text{ s}$. Calculate:
   1. the acceleration of the car
   2. the displacement of the car during the braking.
2. A car is driven at $\scriptsize 25\text{ m}\text{.}{{\text{s}}^{{-1}}}$ in a municipal area. When the driver sees a traffic officer at a speed trap, he realises he is travelling too fast. He immediately applies the brakes of the car while still $\scriptsize 100\text{ m}$ away from the speed trap.
   1. Calculate the magnitude of the minimum acceleration which the car must have to avoid exceeding the speed limit if the municipal speed limit is $\scriptsize 16.6\text{ m}\text{.}{{\text{s}}^{{-1}}}$.
   2. Calculate the time from the instant the driver applied the brakes until he reaches the speed trap. Assume that the car’s velocity, when reaching the trap, is $\scriptsize 16.6\text{ m}\text{.}{{\text{s}}^{{-1}}}$ .
3. A bus on a straight road starts from rest at a bus stop and accelerates at $\scriptsize 2\text{ m}\text{.}{{\text{s}}^{{-2}}}$ until it reaches a speed of $\scriptsize 20\text{ m}\text{.}{{\text{s}}^{{-1}}}$. Then the bus travels for $\scriptsize 20\text{ s}$ at a constant speed until the driver sees the next bus stop in the distance. The driver applies the brakes, stopping the bus in a uniform manner in $\scriptsize 5\text{ s}$.
   1. How long does the bus take to travel from the first bus stop to the second bus stop?
   2. What is the average speed of the bus during the trip?
4. A boy on a bicycle cycles at a constant speed of $\scriptsize 15\text{ m}\text{.}{{\text{s}}^{{-1}}}$ . When he passes the driver of a stationary car, the driver starts to accelerate the car uniformly. If the boy and the driver are in line with one another $\scriptsize 7$ seconds later:
   1. Calculate how far the car has travelled in those $\scriptsize 7$ seconds.
   2. Calculate the acceleration of the car.
   3. Calculate the velocity of the car as it passes the boy.
5. A truck is moving forward at a constant speed of $\scriptsize 20\text{ m}\text{.}{{\text{s}}^{{-1}}}$ on a horizontal road. At point X the driver sees a stationary car ahead and hits the breaks $\scriptsize 0.35\text{ s}$ later at Y. The magnitude of the average acceleration of the truck is $\scriptsize 2.5\text{ m}\text{.}{{\text{s}}^{{-2}}}$. The truck manages to stop $\scriptsize 2\text{ m}$ away from the car at point Z.
   
   1. What is the reaction time of the driver?
   2. Calculate how far the truck travels before the brakes are applied.
   3. Calculate how long it takes before the truck comes to rest from the moment the car is seen.
   4. Calculate the braking distance of the truck.
   5. Write down how far point X is from the car.
6. At the regional championships, Mandla ran the first $\scriptsize 20\text{ m}$ of the $\scriptsize 100\text{ m}$ race in a time of $\scriptsize 3.4\text{ s}$.
   1. Calculate his average speed for the first $\scriptsize 20\text{ m}$ of the race.
   2. For the first $\scriptsize 20\text{ m}$ Mandla’s acceleration is uniform. Calculate the acceleration.
   3. Calculate his speed after the first $\scriptsize 20\text{ m}$ of the race.
   4. Will Mandla be able to better his personal best time of $\scriptsize 10.8\text{ s}$ if he maintains his speed after the first $\scriptsize 20\text{ m}$ of the race? Motivate your answer with a calculation.
7. In $\scriptsize 1938$, a British train, the ‘Mallard’ set a world record for the fastest steam train of $\scriptsize 200\text{ km}\text{.}{{\text{h}}^{{-1}}}$. The train passed Stoke Summit travelling at $\scriptsize 34\text{ m}\text{.}{{\text{s}}^{{-1}}}$. It then accelerated uniformly reaching a maximum speed of $\scriptsize 56\text{ m}\text{.}{{\text{s}}^{{-1}}}$after covering a distance of $\scriptsize 9\text{ 600 m}$.
   1. Convert $\scriptsize 200\text{ km}\text{.}{{\text{h}}^{{-1}}}$ to $\scriptsize \text{m}\text{.}{{\text{s}}^{{-1}}}$.
   2. Calculate the acceleration of the train from the time it passed Stoke Summit to when it reached its maximum speed.
      .
      Once the train had reached its maximum speed it continued at this speed for a further $\scriptsize 500\text{ m}$ when it was decided to stop the train. It took $\scriptsize 120\text{ s}$ to bring the train uniformly to a standstill.
   3. Calculate the distance covered from the moment the train reached maximum speed.

The full solutions are at the end of the unit.

Exercise 2.1

1. Let forward be positive
   $\scriptsize \begin{array}{l}{{v}_{i}}\text{ = 10 m}\text{.}{{\text{s}}^{{-1}}}\\a\text{ = 1 m}\text{.}{{\text{s}}^{{-2}}}\\\Delta t\text{ = 10 s}\\{{v}_{f}}\text{ = ?}\\{{v}_{f}}\text{ = }{{v}_{i}}+a\Delta t\text{ }\\\text{ = 10+1(10)}\\\text{ = 20 m}\text{.}{{\text{s}}^{{-1}}}\text{ forward}\end{array}$
2. Let forward be positive
   $\scriptsize \begin{array}{l}{{v}_{i}}\text{ = 0}\\a\text{ = 1 m}\text{.}{{\text{s}}^{{-2}}}\\\Delta t\text{ = 10 s}\\s\text{ = ?}\\s\text{ = }{{v}_{i}}\Delta t+\displaystyle \frac{1}{2}a\Delta {{t}^{2}}\text{ }\\\text{ = 0 + }\displaystyle \frac{1}{2}(1){{(10)}^{2}}\\\text{ = 50 m }\end{array}$
3. Let forward be positive
   $\scriptsize \begin{array}{l}{{v}_{i}}\text{ = 30 m}\text{.}{{\text{s}}^{{-1}}}\\{{v}_{f}}\text{ = 0}\\\Delta t\text{ = 5 s}\\s\text{ = ?}\\{{v}_{f}}\text{ = }{{v}_{i}}+a\Delta t\text{ }\\\text{0 = 30 + }a(5)\\a\text{ = }-6\text{ m}\text{.}{{\text{s}}^{{-2}}}\\s\text{ = }{{v}_{i}}\Delta t+\displaystyle \frac{1}{2}a\Delta {{t}^{2}}\text{ }\\\text{ = 30(5) + }\displaystyle \frac{1}{2}(-6){{(5)}^{2}}\\\text{ = 75 m forward}\end{array}$
4. Let forward be positive
   $\scriptsize \begin{array}{l}{{v}_{i}}\text{ = 20 m}\text{.}{{\text{s}}^{{-1}}}\\{{v}_{f}}\text{= 0}\\s\text{ = 20 m}\\a\text{ = ?}\\v_{f}^{2}\text{ = }v_{i}^{2}\text{ + 2}as\text{ }\\0\text{ = (20}{{\text{)}}^{2}}\text{ + 2(}a\text{)(20)}\\a\text{ = }-10\text{ m}\text{.}{{\text{s}}^{{-2}}}\\\text{ = 10 m}\text{.}{{\text{s}}^{{-2}}}\text{ backwards}\end{array}$
5. Let forward be positive
   $\scriptsize \begin{array}{l}a\text{ = 4 m}\text{.}{{\text{s}}^{{-2}}}\\{{v}_{i}}\text{ = 0}\\\vartriangle t\text{ = 10 s}\\{{v}_{f}}\text{ = ?}\\s\text{ = ?}\\{{v}_{f}}\text{ = }{{v}_{i}}+a\vartriangle t\text{ }\\\text{ = 0 + 4(10)}\\\text{ = 40 m}\text{.}{{\text{s}}^{{-1}}}\text{ forward}\\s\text{ = }{{v}_{i}}\vartriangle t+\displaystyle \frac{1}{2}a\vartriangle {{t}^{2}}\text{ }\\\text{ = 0 + }\displaystyle \frac{1}{2}(4){{(10)}^{2}}\\\text{ = 200 m forward}\end{array}$
6. Let forward be positive
   $\scriptsize \begin{array}{l}a\text{ = 1}\text{.4 m}\text{.}{{\text{s}}^{{-2}}}\\{{v}_{i}}\text{ = 20 m}\text{.}{{\text{s}}^{{-1}}}\\\Delta t\text{ = 12 s}\\s\text{ = ?}\\{{v}_{f}}\text{ = }{{v}_{i}}+a\Delta t\text{ }\\\text{ = 20 + 1}\text{.4(12)}\\\text{ = 36}\text{.8 m}\text{.}{{\text{s}}^{{-1}}}\text{ forward}\\s\text{ = }{{v}_{i}}\Delta t+\displaystyle \frac{1}{2}a\Delta {{t}^{2}}\text{ }\\\text{ = 20(12) + }\displaystyle \frac{1}{2}(1.4){{(12)}^{2}}\\\text{ = 340}\text{.8 m forward}\end{array}$

Back to Exercise 2.1

Unit 2: Assessment

1. Let east be positive
   1. .
      $\scriptsize \begin{array}{l}{{v}_{i}}\text{ = 30 m}\text{.}{{\text{s}}^{{-1}}}\\{{v}_{f}}\text{ = 20 m}\text{.}{{\text{s}}^{{-1}}}\\\Delta t\text{ = 10 s}\\{{v}_{f}}\text{ = ?}\\a\text{ = ?}\\{{v}_{f}}\text{ = }{{v}_{i}}+a\Delta t\text{ }\\\text{20 = 30 + }a\text{(10)}\\\text{ }a\text{ = }-1\text{ m}\text{.}{{\text{s}}^{{-2}}}\text{ }\\\text{ = }1\text{ m}\text{.}{{\text{s}}^{{-2}}}\text{ west}\end{array}$
   2. .
      $\scriptsize \begin{array}{l}s\text{ = ?}\\s\text{ = }{{v}_{i}}\Delta t+\displaystyle \frac{1}{2}a\Delta {{t}^{2}}\text{ }\\\text{ = 30(10) + }\displaystyle \frac{1}{2}(-1){{(10)}^{2}}\\\text{ = 250 m east}\end{array}$
2. Let forward be positive
   1. .
      $\scriptsize \begin{array}{l}{{v}_{i}}\text{ = 25 m}\text{.}{{\text{s}}^{{-1}}}\\{{v}_{f}}\text{= 16}\text{.6 m}\text{.}{{\text{s}}^{{-1}}}\\s\text{ = 100 m}\\a\text{ = ?}\\\text{ }v_{f}^{2}\text{ = }v_{i}^{2}\text{ + 2}as\text{ }\\\text{16}\text{.}{{\text{6}}^{2}}\text{ = 2}{{\text{5}}^{2}}\text{ + 2}a\text{(10)}\\\text{ }a\text{ = }-1.75\text{ m}\text{.}{{\text{s}}^{{-2}}}\text{ }\\\text{ = }1.75\text{ m}\text{.}{{\text{s}}^{{-2}}}\text{ backwards}\end{array}$
   2. .
      $\scriptsize \begin{array}{l}\text{ }{{v}_{f}}\text{ = }{{v}_{i}}+a\Delta t\text{ }\\16.6\text{ = 25 + (-1}\text{.75)}t\\\text{ }t\text{ = 4}\text{.8 s}\end{array}$
3. .
   1. Let forward be positive
      Find time from A to B
      $\scriptsize \begin{array}{l}{{v}_{i}}\text{ = 0}\\{{v}_{f}}\text{ = 20 m}\text{.}{{\text{s}}^{{-1}}}\\a\text{ = 2 m}\text{.}{{\text{s}}^{{-2}}}\\\Delta t\text{ = ?}\\{{v}_{f}}\text{ = }{{v}_{i}}+a\Delta t\text{ }\\\text{20 = 0 + 2}\Delta t\\\text{ }\Delta \text{t = 10 s }\end{array}$
      .
      $\scriptsize \text{Total time = 1}{{\text{0}}_{{\text{AB}}}}\text{ + 2}{{\text{0}}_{{\text{BC}}}}\text{ + }{{\text{5}}_{{\text{CD}}}}\text{ = 35 s}$
   2. Need to find the total distance from A to D
      $\scriptsize \begin{array}{l}{{s}_{{\text{AB}}}}\text{ = }{{v}_{i}}\Delta t+\displaystyle \frac{1}{2}a\Delta {{t}^{2}}\text{ }\\\text{ = 0 + }\displaystyle \frac{1}{2}\text{(2)(10}{{\text{)}}^{2}}\text{ }\\\text{ = 100 m}\\{{s}_{{\text{BC}}}}\text{ }v\text{ = }\displaystyle \frac{{\Delta s}}{{\Delta t}}\text{ (constant velocity)}\\\text{ 20 = }\displaystyle \frac{{\Delta s}}{{20}}\\\text{ }\Delta {{s}_{{BC}}}\text{ = 400 m}\\{{s}_{{CD}}}\text{ : }{{v}_{i}}\text{ = 20 m}\text{.}{{\text{s}}^{{-1}}}\text{ }{{v}_{f}}\text{ = }{{v}_{i}}+a\Delta t\text{ }\\\text{ }{{v}_{f}}\text{ = 0 0 = 20 + }a\text{(5)}\\\text{ }\Delta \text{t = 5 s }a\text{ = }-4\text{ m}\text{.}{{\text{s}}^{{-2}}}\text{ }\\v_{f}^{2}\text{ = }v_{i}^{2}\text{ + 2}as\text{ }\\\text{ }0\text{ = 2}{{\text{0}}^{2}}\text{ + 2(}-4)(s)\\\text{ }s\text{ = 50 m}\\\text{Total distance = 100 + 400 + 50 = 550 m}\\{{v}_{{av}}}\text{ = }\displaystyle \frac{{{{s}_{{total}}}}}{{{{t}_{{total}}}}}\text{ = }\displaystyle \frac{{550}}{{35}}=\text{ 15}\text{.71 m}\text{.}{{\text{s}}^{{-1}}}\text{ forward}\end{array}$
4. Let forward be positive
   $\scriptsize \begin{array}{l}\text{Boy: }{{v}_{i}}\text{ = 15 m}\text{.}{{\text{s}}^{{-1}}}\text{ (constant velocity) Car: }{{v}_{i}}\text{ = 0 }\\\text{ }\Delta t\text{ = 7 s }\Delta t\text{ = 7 s}\end{array}$
   1. .
      $\scriptsize \begin{array}{l}\Delta {{s}_{{\text{boy}}}}\text{ = }\Delta {{s}_{{\text{car}}}}\text{ (when the car and boy are in line)}\\\text{ }{{v}_{{\text{boy}}}}\text{ = }\displaystyle \frac{{\Delta s}}{{\Delta t}}\\\text{ }15\text{ = }\displaystyle \frac{{\Delta s}}{7}\\\text{ }\Delta {{s}_{{\text{boy}}}}\text{ = }\Delta {{s}_{{\text{car}}}}\text{ = 105 m forward}\end{array}$
   2. .
      $\scriptsize \begin{array}{l}\text{ }s\text{ = }{{v}_{i}}\Delta t+\displaystyle \frac{1}{2}a\Delta {{t}^{2}}\text{ }\\105\text{ = 0 + }\displaystyle \frac{1}{2}a{{(7)}^{2}}\\\text{ }a\text{ = 4}\text{.29 m}\text{.}{{\text{s}}^{{-2}}}\text{ forward}\end{array}$
   3. .
      $\scriptsize \begin{array}{l}{{v}_{f}}\text{ = }{{v}_{i}}+a\Delta t\text{ }\\\text{ = 0 + 4}\text{.29(7)}\\\text{ = 30 m}\text{.}{{\text{s}}^{{-1}}}\text{ forward}\end{array}$
5. .
   1. reaction time $\scriptsize =\text{ 0}\text{.35 s}$
   2. constant velocity during reaction time:
      $\scriptsize \begin{array}{l}\text{ }v\text{ =}\displaystyle \frac{{\Delta s}}{{\Delta t}}\\20\text{ = }\displaystyle \frac{{\Delta s}}{{0.35}}\\\text{ }\Delta s\text{ = 7 m}\end{array}$
   3. Calculate time while braking:
      $\scriptsize \begin{array}{l}{{v}_{i}}\text{ = 20 m}\text{.}{{\text{s}}^{{-1}}}\\{{v}_{f}}\text{= 0}\\a\text{ = }-2.5\text{ m}\text{.}{{\text{s}}^{{-2}}}\text{ (slowing down, therefore negative)}\\\Delta t\text{ = ?}\\{{v}_{f}}\text{ = }{{v}_{i}}+a\Delta t\text{ }\\\text{ }0\text{ = 20 + (}-2.5)\Delta t\\\text{ }\vartriangle t\text{ = 8 s}\\\text{Total time from X to Z = 0}\text{.35 + 8 = 8}\text{.35 s}\end{array}$
   4. .
      $\scriptsize \begin{array}{l}s\text{ = }{{v}_{i}}\Delta t+\displaystyle \frac{1}{2}a\Delta {{t}^{2}}\text{ }\\\text{ = 20(8) + }\displaystyle \frac{1}{2}\text{(}-2.5){{(8)}^{2}}\text{ }\\\text{ = 80 m}\\\end{array}$
   5. .
      $\scriptsize \begin{array}{l}\text{ X to Y = 7 m}\\\text{ Y to Z = 80 m}\\\text{ Z to car = 2 m Total distance = 89 m}\end{array}$
6. Let forward be positive
   1. .
      $\scriptsize \begin{array}{l}s\text{ = 20 m }\\\Delta t=\text{ 3}\text{.4 s}\\{{v}_{{av}}}=\text{ }\displaystyle \frac{{\Delta s}}{{\Delta t}}\text{ }\\\text{ = }\displaystyle \frac{{20}}{{3.4}}\text{ }\\\text{ = 5}\text{.88 m}\text{.}{{\text{s}}^{{-1}}}\text{ }\end{array}$
   2. .
      $\scriptsize \begin{array}{l}{{v}_{i}}\text{ = 0}\\s\text{ = 20 m}\\\Delta t\text{ = 3}\text{.4 s}\\\text{ }s\text{ = }{{v}_{i}}\Delta t+\displaystyle \frac{1}{2}a\Delta {{t}^{2}}\text{ }\\20\text{ = 0 + }\displaystyle \frac{1}{2}a{{(3.4)}^{2}}\\\text{ }a\text{ = 3}\text{.46 m}\text{.}{{\text{s}}^{{-2}}}\text{ forward}\end{array}$
   3. .
      $\scriptsize \begin{array}{l}{{v}_{f}}\text{ = }{{v}_{i}}+a\Delta t\text{ }\\\text{ = 0 + 3}\text{.46(3}\text{.4)}\\\text{ = 11}\text{.76 m}\text{.}{{\text{s}}^{{-1}}}\text{ }\end{array}$
   4. .
      $\scriptsize \begin{array}{l}s\text{ = 80 m}\\v\text{ = 11}\text{.76 m}\text{.}{{\text{s}}^{{-1}}}\\\Delta t\text{ = ?}\\\text{ }v\text{ = }\displaystyle \frac{{\Delta s}}{{\Delta t}}\\11.76\text{ }=\text{ }\displaystyle \frac{{80}}{{\Delta t}}\\\text{ }\Delta t\text{ = 6}\text{.8 s}\\\text{Total }t=\text{ 3}\text{.4 + 6}\text{.8 = 10}\text{.2 s}\\\text{This is better than his personal best of 10}\text{.8 s}\end{array}$
7. .
   1. $\scriptsize 200\text{ }\div \text{ 3}\text{.6 = 55}\text{.56 m}\text{.}{{\text{s}}^{{-1}}}$
   2. Let forward be positive
      $\scriptsize \begin{array}{l}{{v}_{i}}\text{ = 34 m}\text{.}{{\text{s}}^{{-1}}}\\{{v}_{f}}\text{ = 56 m}\text{.}{{\text{s}}^{{-1}}}\\s\text{ = 9 600 m}\\v_{f}^{2}\text{ = }v_{i}^{2}\text{ + 2}as\text{ }\\\text{5}{{\text{6}}^{2}}\text{ = 3}{{\text{4}}^{2}}\text{ + 2}a(9\text{ }600)\\a\text{ = 0}\text{.1 m}\text{.}{{\text{s}}^{{-2}}}\text{ forward}\end{array}$
   3. .
      $\scriptsize \begin{array}{l}{{v}_{i}}\text{ = 56 m}\text{.}{{\text{s}}^{{-1}}}\\{{v}_{f}}=\text{ 0}\\\Delta t\text{ = 120 s}\\s\text{ = ?}\\{{v}_{f}}\text{ = }{{v}_{i}}+a\Delta t\text{ }\\0\text{ = 56 + }a\text{(120)}\\a\text{ = }-0.47\text{ m}\text{.}{{\text{s}}^{{-2}}}\\v_{f}^{2}\text{ = }v_{i}^{2}\text{ + 2}as\text{ }\\\text{ }0\text{ = 5}{{\text{6}}^{2}}\text{ + 2(}-0,47)s\\\text{ }s\text{ = 3 336}\text{.17 m}\\\text{Total distance = 500 + 3 336}\text{.17 = 3 836}\text{.17 m}\end{array}$

Back to Unit 2: Assessment