5 mark questions
12. A car is running at a steady speed of 72 km/h
Its speed is reduced to 18 km/h in just 2 secon
applying brakes. Find (i) its retardation in
time in which the car comes to rest (iii) distance
travelled by the car before coming to rest.
Answers
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2
Answer:
u = 72 kmph
v= 18kmph
t = 2 s = 1 / 1800 hr
i) a = v-u/t = 18 - 72 * 1800/1
= -54 * 1800
= -27000 m/s²
ii) by third equaton of motion,
v² - u² = 2as
ie 18000² - 72000² = 2 * -27000 * s
(skipping the calculation bits),
= 90000 metres
or 9 km.
im not entirely sure, but i think this is right
hope this helped :)
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