Question

The car moving with velocities 10ms^-1and 15ms^-1. Find the ratio of stopping distances if brakes are applied to cause same acceleration

Answer 1

Solution :

⏭ Given:

✏ The car moving with velocities 10mps and 15mps.

⏭ To Find:

✏ Ratio of stopping distance

⏭ Formula:

✏ As per second equation of kinematics

$\mapsto \sf \: {v}^{2} - {u}^{2} = 2as \\ \\ \mapsto \sf \: {0}^{2} - {u}^{2} = 2( - a)s \\ \\ \red{ \sf{ \dag \: negative \: sign \: shows \: retardation}} \\ \\ \mapsto \sf \: - {u}^{2} = - 2as \\ \\ \mapsto \underline{ \boxed{ \bold{ \tt{ \pink{ \large{s = \dfrac{ {u}^{2} }{2a}}}}}}} \: \orange{ \bigstar}$

⏭ Calculation:

✏ Since, acceleration is same for both cases, we can say that...

$\dashrightarrow \sf \: \purple{s \varpropto {u}^{2} } \\ \\ \dashrightarrow \sf \: \dfrac{s_1}{s_2} = \dfrac{ {u_1}^{2} }{ {u_2}^{2} } \\ \\ \dashrightarrow \sf \: \dfrac{s_1}{s_2} = \dfrac{ {10}^{2} }{ {15}^{2} } \\ \\ \dashrightarrow \sf \: \dfrac{s_1}{s_2} = \dfrac{100}{225} \\ \\ \rightarrowtail \: \boxed{ \bold{ \tt{ \large{ \green{s_1 : s_2 = 4 : 9}}}}}$