Answer:
Car fastly moved in 5s is
[tex]{\bf V_{3}} = {\bf 60}\frac{\bf m}{\bf s}[/tex]
Step-by-step explanation:
Given that a car starts from start to rest
so [tex]V_{1} = 0[/tex] and [tex]T_{1}=0[/tex]
and given car reaches a speed of 40 m/s in 10 s.
ie, [tex]V_{2} =40[/tex] and [tex]T_{2} = 10[/tex]
Use the formula
[tex]The rate of change of velocities=acceleration\times the rate of change of times[/tex]
ie, [tex]V_{2} - V_{1} = a(T_{2}-T_{1})[/tex]
[tex]V_{1} = 0[/tex] , [tex]T_{1} = 0[/tex] and a = constant
So, [tex]a = \frac {V_{2}}{T_{2}}[/tex]
[tex]a = \frac {40}{10}[/tex]
[tex]a = 4 \frac {m}{s}^{2}[/tex]
Similarly,
[tex]T_{3} = 15s[/tex] (5 seconds later) and [tex]V_{1} = 0[/tex] , [tex]T_{1} = 0[/tex]
[tex]V_{3} - V_{1} = a\times (T_{3} -T_{1})[/tex]
Then [tex]V_{3}-0= 4\times (15-0)[/tex]
[tex]V_{3} = 4\times (15)[/tex]
[tex]V_{3} = 60\frac {m}{s}[/tex]
