Answered

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A ball is rolled up a ramp with an initial velocity of +6.0m/s. The ball experiences an acceleration of -2.0m/s^2. How can we determine at what time(s) the ball is +5.0m from its initial position on the ramp?

Sagot :

  • Initial velocity=u=6m/s
  • Acceleration=a=-2m/s^2
  • Distance=s=5m
  • Time is t

Apply second equation of kinematics

[tex]\\ \tt\hookrightarrow s=ut+\dfrac{1}{2}at^2[/tex]

[tex]\\ \tt\hookrightarrow 5=6t+\dfrac{1}{2}(-2)t^2[/tex]

[tex]\\ \tt\hookrightarrow 6t-t^2=5[/tex]

[tex]\\ \tt\hookrightarrow t^2-6t+5=0[/tex]

[tex]\\ \tt\hookrightarrow t^2-5t-t+5=0[/tex]

[tex]\\ \tt\hookrightarrow t(t-5)-1(t-5)=0[/tex]

[tex]\\ \tt\hookrightarrow (t-1)(t-5)=0[/tex]

[tex]\\ \tt\hookrightarrow t=(1,5)[/tex]

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