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Kisachek [45]
3 years ago
15

Traveler A starts from rest at a constant acceleration of 6 m/s^2. Two seconds later, traveler B starts with an initial velocity

of 20 m/s at the same acceleration of 6 m/s^2. as measured by a, at what time will traveler B overtake traveler A?
a. 0.4s
b. 1.5s
c. 2.0s
d. 2.5s
e. 3.5s
Physics
1 answer:
Troyanec [42]3 years ago
4 0

Answer:

3. 3.5 s

Explanation:

The position of traveller A is given by the equation:

x_A(t) = \frac{1}{2}a t^2

where

a = 6 m/s^2 is the acceleration of A

t is the time measured from when A started the motion

The position of traveller B instead is given by

x_B(t) = u_B (t-2) + \frac{1}{2}a(t-2)^2

where a (acceleration) is the same as traveller A, and

u_B = 20 m/s

is B's initial velocity. We can verify that the formula is correct by substituting t=2, and we get x_B=0, which means that B starts its motion 2 seconds later.

Traveller B overtakes traveller A when the two positions are the same, so:

x_A = x_B\\\frac{1}{2}at^2 = u_B (t-2) + \frac{1}{2}a(t-2)^2\\\frac{1}{2}at^2 = u_B t - 2u_B +\frac{1}{2}at^2 +2a-2at\\u_Bt-2at = 2u_B-2a\\t=\frac{2u_B-2a}{u_B-2a}=\frac{2(20)-2(6)}{20-2(6)}=3.5 s

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