Respuesta :
Answer:
50 m
Explanation:
Speed of ball during 1st five seconds of motion = v₁
Initial speed of the ball = u
[tex]v_1=u+at\\\Rightarrow v_1=5a[/tex]
Speed of ball during 2nd five seconds of motion = v₂
Initial speed of ball during 2nd five seconds of motion = final Speed of ball during 1st five seconds of motion
[tex]v_2=u+at\\\Rightarrow v_2=5a+5a\\\Rightarrow v_2=10a[/tex]
Distance travelled during 2nd five seconds of motion
[tex]\frac{v_1+v_2}{2}\times 5=150\\\Rightarrow 5a+10a=60\\\Rightarrow a=\frac{60}{15}\\\Rightarrow a=4\ m/s^2[/tex]
Distance travelled during 1st five seconds of motion
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0t+\frac{1}{2}4\times 5^2\\\Rightarrow s=50\ m[/tex]
∴ Distance travelled during 1st five seconds of motion is 50 m
