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A ball starts from rest and rolls down a hill with uniform acceleration, traveling 150 m during the second 5.0 s of its motion. How far did it roll during the first 5.0 s of motion?

1 Answer

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Answer:

50 m

Step-by-step explanation:

Speed of ball during 1st five seconds of motion = v₁

Initial speed of the ball = u


v_1=u+at\\\Rightarrow v_1=5a

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


v_2=u+at\\\Rightarrow v_2=5a+5a\\\Rightarrow v_2=10a

Distance travelled during 2nd five seconds of motion


(v_1+v_2)/(2)* 5=150\\\Rightarrow 5a+10a=60\\\Rightarrow a=(60)/(15)\\\Rightarrow a=4\ m/s^2

Distance travelled during 1st five seconds of motion


s=ut+(1)/(2)at^2\\\Rightarrow s=0t+(1)/(2)4* 5^2\\\Rightarrow s=50\ m

∴ Distance travelled during 1st five seconds of motion is 50 m

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