Asked by jerd
A car starts from a certain point at a velocity of 100 kph. After 20 seconds, another car staring from rest, starts at the same point following car A at a uniform acceleration of 3 m/s². Assuming that the road is boundless will car B overtake car A? If yes, where and when?
Answers
Answered by
Damon
100,000 meters/3600 seconds = 27.8 m/s
d = 27.8 t
d = (1/2)(3)(t-20)^2
so
18.5 t = t^2 -40 t + 400
solve quadratic for t
d = 27.8 t
d = (1/2)(3)(t-20)^2
so
18.5 t = t^2 -40 t + 400
solve quadratic for t
Answered by
Chanz
Sorry Damon. Gives imaginaries.
x = 27.8(t+20)
x = 1/2(3)t^2
1.5t^2=27.8t+556
1.5t^2 - 27.8t - 556 = 0
Real solutions
x = 27.8(t+20)
x = 1/2(3)t^2
1.5t^2=27.8t+556
1.5t^2 - 27.8t - 556 = 0
Real solutions
Answered by
Henry
d = 27.8 * 20 = 556 m. Head
start.
0.5a*t^2 = 556 + 27.8t.
1.5t^2 = 556 + 27.8t.
1.5t^2 - 27.8t - 556 = 0.
t = Time required for B to catchup.
0.5a*t^2 = Distance at which B catches up.
start.
0.5a*t^2 = 556 + 27.8t.
1.5t^2 = 556 + 27.8t.
1.5t^2 - 27.8t - 556 = 0.
t = Time required for B to catchup.
0.5a*t^2 = Distance at which B catches up.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.