A motorcyclist traveling at 30ms\(^{-1}\) starts to apply his brakes when he is 50m from the traffic light that had just turned red. If he reached the traffic light, his decceleration is
18 ms\(^{-2}\)
10 ms\(^{-2}\)
9 ms\(^{-2}\)
5 ms\(^{-2}\)
Explanation
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Discussions (2)
Equation of motion,V²=u² -2aS since they said, He reached the traffic light it means he has come to rest, where we know V =0ms-¹. Let this not confuse you. When a body is decelerating its final velocity reduces until it becomes=0ms-¹ when the body has come to rest. We are having V=0ms-¹ ,u=30ms-¹,a(deceleration)= ?,S(distance) = 50m, applying the formula 0² =30² - 2×a×50
0² = 900 - 100a
-900 ÷ (-100a) = -9ms-² therefore, a = -9ms-². The reason why we are using (-) in that formula (V² =u² - 2aS) is because the motorcyclist is decelerating and the negative answer is showing its direction, ( shows the body is decelerating).
