An auto A is moving at 6m/s and accelerating at 1.5m/s2 to overtake an auto B which is 115.2m ahead. If auto B is moving at 18m/s and decelerating at 0.9m/s2.
a) how soon will A pass B?
b) Determine the distance traveled by B when it overtake B?
Answers
Given : An auto A is moving at 6m/s and accelerating at 1.5m/s2 to overtake an auto B which is 115.2m ahead. auto B is moving at 18m/s and decelerating at 0.9m/s2.
To Find : a) how soon will A pass B?
b) Determine the distance traveled by B when it overtake B?
Solution:
auto A is moving at 6m/s and accelerating at 1.5m/s²
Distance covered in t sec using S= ut + (1/2)at²
= 6t + (1/2)(1.5)t²
auto B is moving at 18m/s and decelerating at 0.9m/s2
= 18t - (1/2)(0.9)t²
auto B is 115.2m ahead.
=> Distance covered by auto A = 115.2 + Distance covered by auto A
=> 6t + (1/2)(1.5)t² = 115.2 + 18t - (1/2)(0.9)t²
=> (1/2)(2.4)t² - 12t - 115.2 = 0
=> 1.2t² - 12t - 115.2 = 0
=> t² - 10t - 96 = 0
=> t² - 16t + 6t - 96 = 0
=> t (t - 16) + 6(t - 16) = 0
=> ( t + 6) (t - 16) = 0
=> t = 16
After 16 secs A will pass B
the distance traveled by B
= 18t - (1/2)(0.9)t²
= 18 * 16 - (1/2)(0.9)(16)²
= 16 ( 18 - 7.2)
= 16 ( 10.8)
= 172.8 m
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