Question

1. two cars A and B travelling in same direction get stopped at a traffic signal. when signal turns green car A accelerates at 0.75 m/s2 and 1.75 sec later car B starts and accelerates at 1.1 m/s2 .Find ( a) when and where B will overtake A and (b) the speed of each car at that time​

Answer 1

Answer:

(a)Car B will overtake Car A after 10.04 sec

(b)Speed of car A=7.53m/s

Speed of Car B=9.11m/s

Explanation:

Car A

Initial velocity is

$u _{A} = 0 \: ms { }^{ - 1}$

Acceleration of car is

$a _{A} = 0.75 \: ms { }^{ - 2} </p><p>$

Car B

Initial velocity is

$u _{B} = 0 \: ms {}^{ - 1}$

Acceleration of car is

$a _{B} = 1.1ms {}^{ - 2}$

Let us assume that,it takes t seconds for Car B to overtake Car A after the start of car A

Given that,Car B starts 1.75 sec after A.Hence,

Running time of B=(t-1.75) sec

Running time of A=(t) sec

Using Kinematic equations of motion for car A and Car B,we get

$S _{A} = u_{A}t + \frac{1}{2} a _{A}t {}^{2} \\ = > S _{A} = \frac{0.75}{2} t {}^{2} = 0.375t {}^{2}$

Also

$S _{B} = u_{B}t + \frac{1}{2} a _{B}t {}^{2} \\ = > S _{B} = \frac{1.1}{2} (t - 1.75) {}^{2} \\ = 0.55(t - 1.75) {}^{2}$

(a)To overtake car A we have condition

$S _{A} = S _{B} \\ = > 0.375t {}^{2} = 0.55(t - 1.75) {}^{2} \\ = > 0.375t {}^{2} = 0.55t {}^{2} - 1.925t + 1.684 \\ = > 0.175t {}^{2} - 1.925t + 1.684 = 0$

Solving the above quadratic equation,we get

$t = 10.04 \: and \: t = 0.95 \\ but \: t > 1.75 \: therefore \\t = 10.04sec$

Hence,Car B will overtake Car A after 10.04sec

(b)Speeds of each car

$V _{A} = 0 + 0.75t = 0.75 \times 10.04 \\ = 7.53ms {}^{ - 1}$

$V _{B} = 0 + 1.1(t - 1.75) \\ = 1.1(10.04 - 1.75) = 9.11ms {}^{ - 1}$

Therefore,the speeds if Car A and Car B are 7.53m/s and 9.11m/s

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