Question

A car moving with a velocity of 20 ms-1 is stopped in a distance of 40 m. If the same car is travelling at double the velocity, the distance
travelled by it for same retardation is?
​

Answer 1

Solution :

⏭ Given:

✏ Velocity of car = 20m/s

✏ Stopping distance = 40m

⏭ To Find:

✏ Stopping distance of the same car if it is travelling at double the velocity.

⏭ Concept:

✏ This question is completely based on concept of stopping distance.

⏭ Formula derivation:

✏ As per third equation of kinematics

$\rightarrow \sf \: {v}^{2} - {u}^{2} = 2as \\ \\ \rightarrow \sf \: {0}^{2} - {u}^{2} = 2( - a)s \\ \\ \rightarrow \sf \: {u}^{2} = 2as \\ \\ \rightarrow \underline{ \boxed{ \pink{ \sf{s = \frac{ {u}^{2} }{2a}}}}} \: \star$

⏭ Calculation:

✏ Here retardation is same for both cases

$\therefore \sf \: \red{ s \propto \: {u}^{2} } \\ \\ \rightarrow \sf \: \frac{s_1}{s_2} = \frac{ {u_1}^{2} }{u_2} \\ \\ \rightarrow \sf \: \frac{40}{s_2} = \frac{ {u_1}^{2} }{( {2u_1)}^{2} } = \frac{1}{4} \\ \\ \rightarrow \sf \: s_2 = 40 \times 4 \\ \\ \rightarrow \: \underline{ \boxed{ \bold{ \sf{ \orange{s_2 = 160 \: m}}}}} \: \red{ \bigstar}$

Answer 2

Explanation:

Initial speed of the car is u.

Final speed of the car v=0

Let the retardation of the  car be a.

Using  v^2−u^2=2aS

Or  0−u2=2aS

⟹ S∝u2   (for same a)

Given :  S=40m    u′=2u

∴ SS′=u2(u′)2=22

Or  40S′=4

⟹  S′=160m