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1. (a) A car has an initial speed of u m s^-1 - Leaving Cert Applied Maths - Question 1 - 2016
Question 1
1. (a) A car has an initial speed of u m s^-1. It moves in a straight line with constant acceleration a for 4 seconds. It travels 40 m while accelerating. The car th... show full transcript
Worked Solution & Example Answer:1. (a) A car has an initial speed of u m s^-1 - Leaving Cert Applied Maths - Question 1 - 2016
Step 1
(i) Draw a speed-time graph for the motion.
Answer
To construct the speed-time graph:
- From the data given, the car accelerates for 4 seconds.
- It reaches a speed of 15 m/s after 4 seconds with constant acceleration.
- The graph starts at (0,0) and increases linearly to the point (4, 15).
- After 4 seconds, the speed remains constant at 15 m/s for 3 seconds (from t = 4s to t = 7s).
- Finally, there is a retarded phase where the speed decreases to 0.
The graph should consist of a trapezoidal shape, illustrating these three phases.
Step 2
(ii) Find the value of u.
Answer
Using the equation of motion:
-
Total distance during acceleration phase (0 to 4 seconds) is given by:
[ s_1 = ut + \frac{1}{2}at^2 ]
40 = 4u + \frac{1}{2}a(4^2)
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The final speed after 4 seconds (v) is:
[ v = u + at ]
15 = u + 4a
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We can solve for a from the second equation:
[ a = \frac{15 - u}{4} ]
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Substituting for a in the distance equation:
[ 40 = 4u + \frac{1}{2} \left(\frac{15 - u}{4}\right)(16) ]
- Simplifying provides:
[ 40 = 4u + \frac{15 - u}{2} ]
- By solving this, we find:
[ u = 5 , m/s ]
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Step 3
(iii) Find the total distance travelled.
Answer
To find the total distance travelled, we sum up the distance from each phase:
-
Distance during acceleration (40 m)
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Distance during uniform speed phase (45 m)
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Distance during deceleration phase:
[ v^2 = u^2 + 2as ]
Here, we can substitute:
[ 0 = 15^2 + 2(-2f)(s) ]
Which can solve for s as well.
Finally,
[ d = 40 + 45 + (3)(15) ]
Total distance:
[ d = 107.5 , m ]
Step 4
(b) Show that h = \( \frac{u^2 - g t^2}{2g} \)
Answer
-
Let the height of the first particle after time t be given by:
[ s_1 = ut - \frac{1}{2}gt^2 ]
- For the second particle after time ( t - \frac{t}{2} = \frac{t}{2} ) the distance becomes:
[ s_2 = u\left(\frac{t}{2}\right) - \frac{g}{2}\left(\frac{t}{2}\right)^2 ]
- When they meet at height h, setting ( s_1 = s_2 ) leads to:
[ ut - \frac{1}{2}gt^2 = \frac{ut}{2} - \frac{g}{8}t^2 ]
- Rearranging gives:
[ h = \frac{u^2 - gt^2}{2g} ]
Thus proving the required expression.