Braking harder and harder before a stop line
Braking harder and harder before a stop line
IP 4
10 marks
A car is travelling at \(15\) m/s when the driver starts to brake, \(30\) m before a stop line. The car's acceleration \(t\) seconds after braking begins is modelled by \(a=-kt\) m/s\(^2\), where \(k\) is a positive constant, until the car stops.
- Find expressions, in terms of \(k\) and \(t\), for the velocity of the car and its distance from the point where braking began.[3]
- Show that the car stops after travelling \(10\sqrt{\dfrac{30}{k}}\) m.[3]
- Find the least value of \(k\) for which the car stops at or before the stop line.[2]
- For this value of \(k\), find the deceleration at the moment the car stops, and comment on whether the model is realistic.[2]
Solution:
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Answer:(a) \(v=15-\dfrac{kt^2}{2}\), \(s=15t-\dfrac{kt^3}{6}\) (b) shown (c) \(k=\dfrac{10}{3}\) (d) \(10\) m/s\(^2\) at \(t=3\), close to the limit of tyre grip, so the model is unrealistic near the stop
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