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A level Mechanics problem, help?
a particle is moving with constant acceleration in a straight line. It passes through three points, A,B and C with speeds 20ms^-1, 30ms^-2 and 45ms^-1 respectively. The time taken to move from A to B is t1 seconds and the time taken to move from B to C is t2 seconds.
a. show that t1/t2 = 2/3
Given also that the total time taken for the particle to move from to C is 50s,
b. find the distance between A and B
Please help! i'm stuck on this one, thanks!
2 Answers
- BittuLv 69 years agoFavorite Answer
a.) Since acceleration is constant, consider it "a"
(final velocity - initial velocity)/time = Accn
a = (30-20)/t1 = 10/t1
a = (45-30)/t2 = 15/t2
10/t1 = 15/t2
Rearranging t1/t2 = 10/15
t1/t2 = 2/3
b.)
If time from A to C is 50s then,
t1 + t2 = 50
t2 = 50 - t1
v = u + at
45 = 20 + (10/t1) * 50 ; constant accn "a" = t1/10
t1 = 20s
Distance between A and B,
s = ut +1/2at^2
s = 20*20 + 1/2 (10/20) * 20^2
s=100 m
- 9 years ago
v(final) - v(initial) = a*t
therefore t = [v(final) - v(initial)]/a
t1 = (30 - 20)/a
t1 = 10/a
t2 = (45 - 30)/a
t2 = 15/a
t1/t2 = [10/a]/[15/a] = (10/a)* (a/15) = 10/15 = 2/3
b)
from ?? to C is 50s
