Question

A car is moving at a speed of 90 km/h. Another car is moving at a speed of 36 km/h and is 300 m behind the first car. If the first car starts to accelerate at 1 m/s^2 and the second car accelerates at 2 m/s^2 then (i) In how much time and (ii) after how much distance will the second car catch up with the first one?​

Answer 1

Answer:

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Answer 2

Given :

Speed of 1st car = 90 km /hr

Speed of other car = 36 km/hr

Distance by which 2nd car is behind the 1st car = 300 m

Acceleration of 1st car = 1 $m/s^2$

Acceleration of 2nd car = 2 $m/s^2$

To Find :

(i) Time taken by the 2nd car to catch the 1st car =?

(ii) After how much distance the 2nd car will catch up the 1st car ?

Solution :

Let the 1st car of car in front be 'A' .

And the 2nd car of car behind A be 'B'

So , the relative velocity of car B w.r.t car A is :

=$V_B - V_A$

= (36-90) km /hr

=-54 km /hr

= $-54 \times \frac{5}{18}$ m/s

= -15 m/s

Relative acceleration of B w.r.t A :

=$a_B - a_A$

= (2-1) = 1  $m/s^2$

Now using 2nd equation of motion i.e. :

$S = ut + \frac{1}{2}at^2$

Or,$300 = -15 t + \frac{1}{2}\times 1 \times t^2$

Or,$t^2 - 30 t -600 = 0$

Or , $t = \frac{30 \pm \sqrt {(30)^2 - 4 \times 1 \times (-600)}}{2}$

         =$\frac{30 \pm \sqrt {3300}}{2}$

         =$\frac{30 \pm 57.44}{2}$

         =$\frac{30 + 57.44}{2}$

         =43 .72 s

We will consider the negative time .

Now for the 2nd part :

Distance travelled  y B :

Since speed of B is = 36 km/hr

So we convert it into m/s , so speed of B will be :

=$36 \times \frac{5}{18}$ m/s

= 10 m/s

$S' = 10 \times 43.72 + \frac{1}{2} \times 2 \times (43.72)^2$

   = 437.2 + 1911.43 m

   =2348.63 m

Hence , the time taken for car B to catch A is 43.72 s and the distance travelled by car B in catching A is 2348.63 m .