Answer:
Initial speed of
car /truck, \[\upsilon =72\text{ }km/h\]
\[=72\,\,\times
\,\frac{5}{18}\,\,=\,20\,m{{s}^{-1}}\]
Acceleration of car, \[{{a}_{c}}=-\,\frac{20}{3}\,m{{s}^{-2}}\,\]
Acceleration of truck, \[{{a}_{t}}=\,-\frac{20}{5}\,=-4\,m{{s}^{-2}}\]
Speed of truck after time \[t,\text{
}{{u}_{t}}=u+{{a}_{t}}t\]
\[=204t~\] ?? (i)
The car will retard after
time \[0.5\,s\]
\ speed of car after time
\[{{(t0.5)}_{s}},\text{
}{{\upsilon }_{c}}=\upsilon +{{a}_{t}}t\]
\[=20\,-\frac{20}{3}\,(t-0.5)\] ?.. (ii)
From eqns. (i) and (ii)
\[20\,-4t=20-\frac{20}{3}\,(t-0.5)\]
or \[t=1.25\,s.\]
Distance travelled by truck
in time \[t=1.25\,s,\]
\[{{S}_{t}}=\,ut+\frac{1}{2}\,a{{t}^{2}}=20\,\times
1.25\,-\,\frac{1}{2}\,\times \,4\,\times \,{{(1.25)}^{2}}\]
\[=21.88\text{ }m.~\]
Distance travelled by car
in time 0.5 s (before it starts retarding), \[{{S}_{1}}=ut=20\times
0.5=10\text{ }m\].
Distance travelled by car
in time \[(t-0.5)s\]
\[=(1.25-0.5)s\] or \[0.75\text{
}s,\]
\[{{S}_{2}}=\,ut+\frac{1}{2}\,a{{t}^{2}}=\,20\,\times
\,0.75\,-\frac{1}{2}\,\,\times \,\frac{20}{3}\,\times \,{{(0.75)}^{2}}\]
\[=151.875=13.125\text{
}m~\]
\ Total distance travelled by car,
\[{{S}_{c}}={{S}_{1}}+{{S}_{2}}=23.125\text{
}m\]
Relative separation between
truck and car to avoid collision
\[={{S}_{c}}-{{S}_{t}}=23.125-21.88=1.25m\]
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