Question

A bus was moving in a speed of 5 km / h on applying brakes it stopped in 8 seconds . Calculate the acceleration and the distance traveled before starting.

Answer 1

Solution:

⏭ Given:

✏ Initial velocity of bus = 5kmph

✏ Final velocity of bus = 0

✏ Time interval = 8s

⏭ To Find:

• Acceleration
• Distance covered by bus

⏭ Concept:

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

⏭ Formula:

• Stopping time

$\star \: \underline{ \boxed{ \bold{ \sf{ \pink{ \large{t = \frac{u}{a}}}}}}} \: \star$

• Stopping distance

$\star \: \underline{ \boxed{ \bold{ \sf{ \purple{ \large{d = \dfrac{ {u}^{2} }{2a}}}}}}} \: \star$

⏭ Terms indication:

• u denotes initial velocity
• a denotes retardation
• t denotes time interval
• d denotes distance

⏭ Conversation:

• 5kmph = 1.39mps

⏭ Calculation:

____________________________

• Acceleration

$\implies \sf \: 8 = \dfrac{1.39}{a} \\ \\ \implies \sf \: a = \dfrac{1.39}{8} \\ \\ \implies \: \boxed{ \sf{ \red{a = 0.17 \: m {s}^{ - 2} }}} \: \gray{ \surd}$

____________________________

• Distance

$\implies \sf \: d = \dfrac{ ({1.39)}^{2} }{2 \times 0.17} \\ \\ \implies \sf \: d = \dfrac{1.93}{0.34} \\ \\ \implies \: \boxed{ \sf{ \blue{d = 5.68 \: m}}} \: \gray{ \surd}$

____________________________

Answer 2

Answer:

-0.1736m/s^2

16.66m

Explanation:

u=initial velocity=5km/h

5 km/h =5*5/18=25/18 m/s

v=final velocity=0 (as the bus will stop)

t=time=8 seconds

first equation of motion says v=u+at

so,

0=25/18+8a

-25/18=8a

a=-25/18×1/8

a= -25/144 = -0.1736m/s^2

-0.1736m/s^2

s=ut+1/2at^2

s=25/18×8+1/2×-0.1736×8×8

100/9+5.55

11.11+5.55

=16.66