A car moving at 160 km/h when passes the mark-A, driver applies brake and reduces its speed uniformly to 40 km/h at mark-C. The marks are spaced at equal distances along the road as shown below. At which part of the track the car has instantaneous speed of 100 km/h? Neglect the size of the car.

Question

A car moving at 160 km/h when passes the mark-A, driver applies brake and reduces its speed uniformly to 40 km/h at mark-C. The marks are spaced at equal distances along the road as shown below. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/78df7c6d90bf7fcc015865df5db00786-.png" /> At which part of the track the car has instantaneous speed of 100 km/h? Neglect the size of the car. (A) At mark- B (B) Between mark-A and mark-B (C) between mark-B and mark-C (D) insufficient information to decide

Tardigrade Admin · Accepted Answer

From A to C: (40)2 - (160)2 = 2a(AC) From A to D: (100)2 - (160)2 = 2a(AD) Here D is the point where speed is 100 km/h. From above AD = (13/20) (AC) and AB = (AC/2) We see that AD > AB.

Q. A car moving at 160km/h when passes the mark-A, driver applies brake and reduces its speed uniformly to 40km/h at mark-C. The marks are spaced at equal distances along the road as shown below. At which part of the track the car has instantaneous speed of 100km/h? Neglect the size of the car.

At mark- B

Between mark-A and mark-B

between mark-B and mark-C

insufficient information to decide

From A to C:(40)2−(160)2=2a(AC) From A to D:(100)2−(160)2=2a(AD) Here D is the point where speed is 100km/h. From above AD=2013​(AC) and AB=2AC​ We see that AD>AB.

From $A$ to $C : (40)^{2} - (160)^{2} = 2 a (A C)$

From $A$ to $D : (100)^{2} - (160)^{2} = 2 a (A D)$

Here $D$ is the point where speed is $100 km / h$ .

From above $A D = \frac{13}{20} (A C)$ and $A B = \frac{A C}{2}$

We see that $A D > A B$ .