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Ludmilka [50]
3 years ago
13

Cars A and B travel along the same straight track. Car A is located at position s1 at clock reading t1 and it maintains a consta

nt speed of v1. Car B is located at position s2 < s1 at clock reading t2, and it maintains a constant speed of v2 > v1. Find an expression for the clock reading at which car B will overtake car A. Find also an expression for the position at which car B overtakes car A. Evaluate your expressions for the values s1 = 0.75 meters, t1 = 0.00 seconds, v1 = 0.38 m s−1 , s2 = 0.0 meters, t2 = 0.08 seconds, and v2 = 0.78 m s−1 .
Physics
1 answer:
m_a_m_a [10]3 years ago
7 0

Answer:

a) t=\frac{S1-V1*t1 - S2 + V2*t2}{V2-V1}

b) X = S2 - V2*t2 + V2*\frac{S1-V1*t1 - S2 + V2*t2}{V2-V1}

c) t = 2.031s      X = 1.52m

Explanation:

Let X1 be position of car A as a function of time and X2 be the position of car B as a function of time:

X1 = Xo1 + V1*t     If we evaluate this expression at t=t1

S1 = Xo1 + V1*t1    Solving for Xo1 we get its initial position at t=0:

Xo1 = S1 - V1*t1    

Similarly for car B:

Xo2 = S2 - V2*t2

Now, to find the instant when car B overtakes car A:

X1 = X2   Replacing their functions in time:

(S1 - V1*t1) + V1*t  = (S2 - V2*t2) + V2*t   Solving for t:

t=\frac{S1-V1*t1 - S2 + V2*t2}{V2-V1}   This is the instant when B reaches A

If we replace this expression in either X1 or X2 expression, we get that position:  X1 = X2

X1 = X2 = X = S2 - V2*t2 + V2*\frac{S1-V1*t1 - S2 + V2*t2}{V2-V1}

If we evaluate these two expressions with the given values:

t = 2.031s  and   X = 1.52m

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6.71 *10^{-5} rad

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A bus slows with constant acceleration from 24.0 m/s to 16.0 m/s and moves 50.0 m in the process. (a) How much further does it t
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Answer:

(a) Bus will traveled further a distance of 40 m

(b) It will take 7.5 sec to stop the bus

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We have given initial velocity of the bus u = 24 m/sec

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Distance traveled in this process s = 50 m

From third equation of motion we know that v^2=u^2+2as

16^2=24^2+2\times a\times 50

a=-3.2m/sec^2

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0^2=24^2-2\times 3.2\times s

s= 90 m

So further distance traveled by bus = 90-50 =40 m

(b) Now as the bus finally stops so final velocity v= 0 m/sec

Initial velocity u = 24 m/sec

Acceleration a=-3.2m/sec^2

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Can work done=mass*acceleration*displacement(work=m*a*s)
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no, work is = force * distance or displacement


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Two vectors are presented as a=3.0i +5.0j and b=2.0i+4.0j find (a) a x b, ab (c) (a+b)b and (d) the component of a along the dir
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Let's ask this question step by step:
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 Part (c)
 (a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
 (a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
 (a + b) b = 10 + 36
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 Part (d)
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 answer
 2k
 26
 46
 26 / root (20)
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2nd derivative gives acceleration;
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For a given time, like 2 seconds, t will be 2. And answer of speed will be scalar.
6 0
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