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A motorcyclist is traveling along a straight road with constant acceleration, a m s-2 The motorcyclist passes point A with speed Ukm/h. After 8 seconds it passes point B, where AB = 50 m. After a further 12 seconds it passes point C, where BC = 250 m.

Find the value of a giving your answer as an exact fraction. [3 marks]

Find U. [2 marks]​

Respuesta :

The times taken and distances covered in AB and BC are combined to

form the second of two equations used to find a and U.

Correct responses:

  • [tex]\displaystyle \underline{ a = 1\frac{11}{24} \ m\cdot s^{-2}}[/tex]
  • U = 1.5 km/h

Methods by which the above responses are derived:

Acceleration of the motorcycle = a m/s²

Speed with which the motorcyclist passes point A = U km/h

Time it takes to travel distance AB = 50 m. is 8 seconds

Time it takes to travel distance BC = 250 m is 12 seconds

Required:

The value of the constant acceleration, a

Solution:

The kinematic equation of motion, that gives the distance travelled, s, is presented as follows;

  • [tex]\displaystyle s = \mathbf{ u \cdot t + \frac{1}{2} \cdot a \cdot t^2}[/tex]

Therefore;

We have;

[tex]\displaystyle 50 = u \times 8 + \frac{1}{2} \times a \times 8^2= 8\cdot u + 32 \cdot a[/tex]

50 = 8·u + 32·a...(1)

After a further 12 seconds, the total time = 12 s + 8 s = 20 s

The total distance = 50 m + 250 m = 300 m

Which gives;

[tex]\displaystyle 300 = 20 \cdot u + \frac{1}{2} \cdot a \cdot 20^2 = 20 \cdot u + 200 \cdot a[/tex]

300 = 20·u + 200·a...(2)

Multiplying equation (1) by 5 and equation (2) by 2 gives;

50 × 5 = 5 × (8·u + 32·a) = 40·u + 160·a

250 = 40·u + 160·a...(3)

300 × 2 = 2 × (20·u + 200·a) = 40·u + 400·a

600 = 40·u + 400·a..(4)

Subtracting equation (3) from equation (4) gives;

600 - 250 = 40·u - 40·u + 400·a - 160·a = 240·a

350 = 240·a

[tex]\displaystyle a = \frac{600 - 250}{240} = \mathbf{\frac{35}{24}}[/tex]

  • [tex]\displaystyle \underline{ a = \frac{35}{24} \, m/s^2 = 1\frac{11}{24} \, m/s^2}[/tex]

From equation (1), we have;

[tex]\displaystyle 50 = \mathbf{ u \cdot 8 + 32 \times \frac{35}{24}}[/tex]

Therefore;

[tex]\displaystyle u = \frac{50 - 32 \times \frac{35}{24} }{8} = \mathbf{ \frac{5}{12}}[/tex]

[tex]\displaystyle The \ initial \ velocity, \ u = \mathbf{ \frac{5}{12} \, m/s}[/tex]

The initial velocity in km/h is therefore;

From equation (1), we have;

[tex]\displaystyle 50 = 8 \cdot u + 32 \times \frac{175}{24}[/tex]

  • [tex]\displaystyle U = \frac{5}{12} \, m/s \times \frac{1 \, km}{1000 \, m} \times \frac{3600 \, s}{1 \, hr} = \frac{3}{2} \, km/h = \underline{ 1.5 \, km/h}[/tex]

Learn more about velocity and acceleration here:

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