Question

If train is at 90 km per hour breaks are applied and the acceleration is -0 .5 then find distance covered before going to rest

Answer 1

${ \red{ \huge{ \mathfrak{ \underline{ \underline{Question:-}}}}}}$

▪IF train is at 90 km per hour breaks are applied and the acceleration is -0.5m/sec^2 . then , find the distance before going to rest??

${ \red{ \huge{ \underline{ \underline{ \mathfrak{Solution:-}}}}}}$

${ \orange{ \bold{ \underline{Given \: that-}}}}$

• initially the train is moving at speed 90km/hr

I.e., initial velocity ( u ) = 90km/hr

${ \bold{ = 90 \times \frac{5}{18} m \: {sec}^{ - 1} = 5 \times 5 m \: {sec}^{ - 1} }}$

${ \pink{ \bold{ initial \: velocity(u) = 25 \: m \: {sec}^{ - 1}}}}$

${ \pink{ \bold{ acceleration(a) = - 0.5 \: m \: {sec}^{ - 2} }}}$

• finally the train comes at rest....

therefore,

${ \bold{ \pink{final \: velocity (v)= 0 \: m \: {sec}^{ - 1}}}}$

${ \orange{ \bold{ \underline{To \: find-}}}}$

▪ The distance(S) covered by the train before going to rest???

Using THIRD EQUATION OF MOTION ,,

for calculating the distance covered by the train before going to rest ......

${ \boxed{ \bold{ \red{ \: \: {v}^{2} = {u}^{2} + 2aS\: \: \: }}}}$

where,

▪ v = final velocity

▪ u = initial velocity

▪ a = acceleration

▪ S = displacement

here, solving for S.....

${ \red{ \bold{ {(0m \: {sec}^{ - 1}) }^{2} = {(25m \: {sec}^{ - 1}) }^{2} + 2 \times ( - 0.5m \: {sec}^{ - 2} ) \times S }}}$

${ \bold{ \implies{0 \: {m}^{2} {sec}^{ - 2} = 625 {m}^{2} \: {sec}^{ - 2} - 1.0 \: m \: {sec}^{ - 2} \times S }}}$

${ \bold{ \implies{1.0 \: m \: {sec}^{ - 2} \times S = 625 \: {m}^{2} \: {sec}^{ - 2} - 0 {m}^{2} \: {sec}^{ - 2} }}}$

${ \bold{ \implies{1.0m \: {sec}^{ - 2} \times S= 625 {m}^{2} \: {sec}^{ - 2}}}}$

${ \bold{ \implies{S = \frac{625 {m}^{2} \: {sec}^{ - 2} }{1m \: {sec}^{ - 2} }}}}$

${ \boxed{ \implies{ \bold{ \red{S = 625 \: m \: \: }}}}}$

therefore,

the distance covered by the train before going to rest is 625 m.....

Answer 2

${ \red{ \huge{ \mathfrak{ \underline{ \underline{Question:-}}}}}}$

▪IF train is at 90 km per hour breaks are applied and the acceleration is -0.5m/sec^2 . then , find the distance before going to rest??

${ \red{ \huge{ \underline{ \underline{ \mathfrak{Solution:-}}}}}}$

${ \orange{ \bold{ \underline{Given \: that-}}}}$

• initially the train is moving at speed 90km/hr

I.e., initial velocity ( u ) = 90km/hr

${ \bold{ = 90 \times \frac{5}{18} m \: {sec}^{ - 1} = 5 \times 5 m \: {sec}^{ - 1} }}$

${ \pink{ \bold{ initial \: velocity(u) = 25 \: m \: {sec}^{ - 1}}}}$

${ \pink{ \bold{ acceleration(a) = - 0.5 \: m \: {sec}^{ - 2} }}}$

• finally the train comes at rest....

therefore,

${ \bold{ \pink{final \: velocity (v)= 0 \: m \: {sec}^{ - 1}}}}$

${ \orange{ \bold{ \underline{To \: find-}}}}$

▪ The distance(S) covered by the train before going to rest???

Using THIRD EQUATION OF MOTION ,,

for calculating the distance covered by the train before going to rest ......

${ \boxed{ \bold{ \red{ \: \: {v}^{2} = {u}^{2} + 2aS\: \: \: }}}}$

where,

▪ v = final velocity

▪ u = initial velocity

▪ a = acceleration

▪ S = displacement

here, solving for S.....

${ \red{ \bold{ {(0m \: {sec}^{ - 1}) }^{2} = {(25m \: {sec}^{ - 1}) }^{2} + 2 \times ( - 0.5m \: {sec}^{ - 2} ) \times S }}}$

${ \bold{ \implies{0 \: {m}^{2} {sec}^{ - 2} = 625 {m}^{2} \: {sec}^{ - 2} - 1.0 \: m \: {sec}^{ - 2} \times S }}}$

${ \bold{ \implies{1.0 \: m \: {sec}^{ - 2} \times S = 625 \: {m}^{2} \: {sec}^{ - 2} - 0 {m}^{2} \: {sec}^{ - 2} }}}$

${ \bold{ \implies{1.0m \: {sec}^{ - 2} \times S= 625 {m}^{2} \: {sec}^{ - 2}}}}$

${ \bold{ \implies{S = \frac{625 {m}^{2} \: {sec}^{ - 2} }{1m \: {sec}^{ - 2} }}}}$

${ \boxed{ \implies{ \bold{ \red{S = 625 \: m \: \: }}}}}$

therefore,

the distance covered by the train before going to rest is 625 m.....