Answer :
Answer:
66.85 m
Explanation:
We are given that
Acceleration ,a=[tex]2.7m/s^2[/tex]
Speed of truck, v=9.5 m/s
We have to find the distance beyond which the traffic signal will the automobile overtake the truck.
Initial speed of automobile, u=0
We know that
[tex]s=ut+\frac{1}{2}at^2[/tex]
Using the formula
[tex]s=0+\frac{1}{2}(27)t^2=\frac{27}{2}t^2[/tex]
For constant speed
Acceleration, a=0
Again
[tex]s=vt+0=9.5t[/tex]
[tex]9.5t=\frac{27}{2}t^2[/tex]
[tex]t=\frac{9.5\times 2}{2.7}=7.037s[/tex]
Substitute the value of t
[tex]x=9.5(7.037)=66.85m[/tex]
Hence, the distance beyond which the traffic signal will the automobile overtake the truck=66.85 m