Speed of car as well as truck = 72 km/h = 72 × \(\frac{5}{18}\) m/s = 20 m/s
Time required to stop the truck = 5 s.
Car behind the truck
Retardation of truck = \(\frac{20}{5}\) = 4 ms-2
Retardation of car = \(\frac{20}{3}\) ms-2
Let the truck be at a distance x from the car when brakes are applied.
Distance of truck from A at t > 0.5 s is x + 20t – 22.
Distance of car from A is
10 + 20(t – 0.5) − \(\frac{10}{3}\) ( t − 0.5)2.
If they meet
x + 20t – 2t2 = 10 + 20t – 10 – \(\frac{10}{3}\)t2 + \(\frac{10}{3}\)t − 0.25 × \(\frac{10}{3}\).
x = − \(\frac{4}{3}\)t2 + \(\frac{10}{3}\)t − \(\frac{5}{6}\).
To find xmin
\(\frac{dx}{dt}=-\frac{8}{3}t+\frac{10}{3}= 0\)
which gives tmin = \(\frac{10}{8}\) = \(\frac{5}{4}\)s .
Therefore, xmin = − \(\frac{4}{3}{(\frac{5}{4})}^2+\frac{10}{3}\times\frac{5}{4}-\frac{5}{6}=\frac{5}{4}\).
Therefore, x > 1.25 m.
Second method : This method does not require the use of calculus.
If the car is behind the truck,
vcar = 20 −(\(\frac{20}{3}\))(t − 0.5) for t > 0.5 s as car decelerate only after 0.5 s.
vtruck = 20 – 4t
Find t from equating the two or from velocity vs tine graph. This yield t = \(\frac{5}{4}\) s.
In this time truck would travel truck,
struck = 20(\(\frac{5}{4}\)) – (\(\frac{1}{2}\))(4)(\(\frac{5}{4}\))2 = 21.875 m
and car would travel.
scar = 20(0.5) + 20( \(\frac{5}{4}\) − 0.5).− ( \(\frac{1}{2}\) ) (\(\frac{20}{3}\)) × ( \(\frac{5}{4}\) − 0.5)2 = 23.125 m.
Thus scar − struck = 1.25 m.
If the car maintains this distance initially, its speed after 1.25 s will be always less than that of truck and hence collision never occurs.
