Question

a bus is moving with a speed of 72 km per hour can be stop by break after at least 10 m. what will be the maximum stopping distance, if the same bus is moving at a speed of 144 km per hour? solve step by step please

Answer 1

u=72km/h
=72×5/18
=20m/s
v=0
S=10m
SINCE.....
${v}^{2} - {u}^{2} = 2as \\ 0 - {20}^{2} = 2a \times 10 \\ - 400 = 20a \\ \frac{ - 400}{20} = a \\ - 20 \frac{m}{ {s}^{2} } = a$


so acceleration is -20m/s^2
NOW,
u=144km/h
=144×5/18
=40m/s
v=0
a= -20m/s^2
S=?
AGAIN,
${v}^{2} - {u}^{2} = 2as \\ 0 - {40}^{2} = 2 \times ( - 20) \times s \\ - 1600 = - 40s \\ \frac{ - 1600}{ - 40} = s \\ 40m = s$

hence maximum stopping s
distance if bus moves at speed of 144km/h
=40m

Answer 2

Answer:

Explanation:

u=72km/h

=72×5/18

=20m/s

v=0

S=10m

SINCE.....

{v}^{2} - {u}^{2} = 2as \\ 0 - {20}^{2} = 2a \times 10 \\ - 400 = 20a \\ \frac{ - 400}{20} = a \\ - 20 \frac{m}{ {s}^{2} } = a

so acceleration is -20m/s^2

NOW,

u=144km/h

=144×5/18

=40m/s

v=0

a= -20m/s^2

S=?

AGAIN,

{v}^{2} - {u}^{2} = 2as \\ 0 - {40}^{2} = 2 \times ( - 20) \times s \\ - 1600 = - 40s \\ \frac{ - 1600}{ - 40} = s \\ 40m = s