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Standard XII

Mathematics

Definition of Vector

A car A is tr...

Question

A car $A$ is travelling on a straight level road with a uniform speed of $60 k m p h$ It is followed by another car $B$ which is moving with a speed of $70 k m p h$ . When the distance between them is $2.5 k m$ , the car $B$ is given a deceleration of $20 k m / h^{2}$ . After how much time will $B$ catch up with $A$

1 h r

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\frac{1}{2} h r

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\frac{1}{4} h r

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\frac{1}{8} h r

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Solution

The correct option is B $\frac{1}{2} h r$
Solving in the rest frame of A

$u_{B A} = u_{B} - u_{A} = 70 - 60 = 10 k m p h a_{B A} = a_{B} - a_{A} = - 20 - 0 = - 20 k m p h^{2} (B travels 2.5 km to meet A) S_{B A} = 2.5 k m S_{B A} = u_{B A} t + \frac{1}{2} a_{B A} t^{2} 2.5 = 10 t - 10 t^{2} \Rightarrow t^{2} - t + 0.25 = 0 t = \frac{1 \pm \sqrt{1 - 1}}{2} t = \frac{1}{2} h r .$

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A car A is travelling on a straight level road with a uniform speed of 60 kmph It is followed by another car B which is moving with a speed of 70 kmph. When the distance between them is 2.5 km, the car B is given a deceleration of 20 km/h2. After how much time will B catch up with A

The correct option is B 12 hrSolving in the rest frame of A uBA=uB−uA=70−60=10 kmphaBA=aB−aA=−20−0=−20 kmph2(B travels 2.5 km to meet A) SBA=2.5 kmSBA=uBAt+12aBAt22.5=10t−10t2⇒t2−t+0.25=0t=1±√1−12t=12 hr.