Question

Ramesh can bring a car at 30km/h to a halt by applying brakes with in a distance of 10m what distance would the same car cover before coming to rest if it's initial velocity be 60km/h​

Answer 1

$\large{\red{\bold{\underline{\underline{Answer}}}}}$

$\purple{\sf{\mapsto 40.52 \: m}}$

$\bold{\green{\underline{Given}}} \\ \\ \sf{\rightsquigarrow Initial \: velocity \: (u) = 30 \: Km/h} \\ \\ \bold{\blue{\underline{To \: Find}}} \\ \\ \sf{\rightsquigarrow Distance \: (s) \:=\:?}$

$\rule{195}{3}$

• u denoted by Initial velocity
• v denoted by Final velocity
• s denoted by Distance

$\rule{195}{3}$

⠀⠀

❏ According To Given Question

⠀⠀

$\bf{\underline{We \: know \: that }} \\ \\ \sf{ : \implies {v}^{2} - {u}^{2} = 2as} \\ \\ \sf{ : \implies v =o \: \: \: \: (Final \: velocity)} \\ \\ \sf{: \implies u \longrightarrow Initial \: velocity,} \\ \\ \sf{ : \implies u = 30 \: Km/hr} \\ \\ \sf{: \implies Km/hr \longrightarrow m/s} \\ \\ \sf{: \implies 30 \times \frac{5}{18} \:m/s } \\ \\ \sf{: \implies 8.3 \: m/s} \\ \\ \bf{On \: applying} \\ \\ \sf{: \implies 0 - {(8.3)}^{2} = - 2 \times a \times 10} \\ \\ \sf{: \implies a = 3.4 \: m/s} \\ \\ \bf{Now,} \\ \\ \sf{: \implies u = 60 \: km/hr} \\ \\ \sf{ : \implies In \: m/s \:\: 60 \times \frac{5}{18} } \\ \\ \sf{: \implies 16.6 \: m/s} \\ \\ \bf{ \: \: \: \: a = 3.4 \: m/s \: \: \: \: \: \: \: \: \: s = ? } \\ \\ \sf{ : \implies 0 - {(16.6)}^{2} = - 2 \times 3.4 \times s } \\ \\ \sf{ \purple{ : \implies \underline{ \boxed{ \sf{ \purple{40.52 \: m}}}}}}\:\:\:\:\:\sf{\orange{(Approx)}}$

Answer 2

$\rule{200}3$

$\sf\pink{Solution}$

• Initial velocity, u = 30 km/h

• Final velocity, v = 0 m/s

• Distance coverd (s) = 10 m

Convert the unit of u,

$\implies$$\:$ $\sf 30 \times \dfrac{5}{18}$

$\implies$ $\:$ $\sf 10 \times \dfrac{5}{6}$

$\implies$ $\:$ $\sf \dfrac{50}{6}$

$\implies$ $\:$ $\sf 8.33 ms^{-1}$

$\therefore$ Initial velocity (u) = 8.33 m/s.

Now, use the third equation of motion,

$\leadsto$ $\:$ $\sf v^{2} = u^{2} + 2as$

$\leadsto$ $\:$ $\sf 0^{2} = \left (8.33 \right)^{2} + 2a(10)$

$\leadsto$ $\:$ $\sf 0 = 69.38 + 20a$

$\leadsto$ $\:$ $\sf -69.38 = 20a$

$\leadsto$ $\:$ $\sf a = \dfrac{-69.38}{20}$

$\leadsto$ $\:$ $\sf a = -3.47 ms^{-2}$

We know that the value of a. Also we have the value of initial velocity (u) for the second case, using the same equation we can here easily solve for distance covered.

We know,

• u = 60 km/h

• a = -3.47 m/s²

• v = 0 m/s

Changing the unit of u,

$\mapsto$ $\:$ $\sf 60 \times \dfrac{5}{18}$

$\mapsto$ $\:$ $\sf \dfrac{300}{18}$

$\mapsto$ $\:$ $\sf 16.67$

$\therefore$ u = 16.67 m/s

What we are going to solve for is the distance covered denoted by (s).

$\dashrightarrow$ $\:$ $\sf v^{2} = u^{2} + 2as$

$\dashrightarrow$ $\:$ $\sf 0^{2} = \left(16.67\right)^{2} + 2\left(-3.47\right) s$

$\dashrightarrow$ $\:$ $\sf 0 = 277.88 + \left(-6.94\right) s$

$\dashrightarrow$ $\:$ $\sf -277.88 = -6.94 s$

$\dashrightarrow$ $\:$ $\sf \dfrac{-277.88}{-6.94} = s$

$\dashrightarrow$ $\:$ $\sf \dfrac{277.88}{6.94} = s$

$\pink\dashrightarrow$ $\:$ $\large\underline{\boxed{\pink{\sf 40.04}}}$ $\dagger$

$\therefore$ distance Ramesh would cover before coming to rest traveling with same car would be 40 metre (approx).

$\rule{200}3$