A motorist travelling at 10 ms-1 can bring his car to rest in a braking distance of 10 m. In what distance could he bring the car to rest from the period of 30 ms-1 using the same breaking force?
Correct answer: D. 90 m
- A. 17m
- B. 30 m
- C. 52 m
- D. 90 m
Explanation
Initial velocity = u= 10 m/s Final velocity = v = 0 m/s Distance covered = 10 m Acceleration of the bike: 2as= v²- u²2(a)(10) = 0² - 10²20a = -100 a= -100/20 a = -5 m/Therefore in the next condition,Initial velocity = u= 30 m/s Final velocity = v = 0 m/sAcceleration = -5 m/s² (as deduced)Therefore distance travelled by the bike, 2as= v²- u²2(-5)(s)= 0² - 30²-10s = -900 s= -900/-10 s= 90 m
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