Question

bus decrease its speed from 90kmph to 54kmph , in 4s. find acceleration of bus
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Answer 1

Answer

Acceleration of the bus = - 2.5 m/s²

Given

Initial speed of the bus is 90 kmph

Final speed of the bus is 54 kmph

Time taken for is 4 s

To Find

Acceleration of the bus

Solution

Initial velocity , u = 90 km/h = 25 m/s

Final velocity , v = = 54 km/h = 15 m/s

Time , t = 4 s

Acceleration , a = ? m/s²

Now , " Acceleration is defined as rate of change in velocity "

$\implies \bf a=\dfrac{v-u}{t}\\\\\implies \rm a=\dfrac{15-25}{4}\\\\\implies \rm a=\dfrac{-10}{4}\\\\\implies \bf a=-2.5\ m/s^2\ \pink{\bigstar}$

Acceleration of the bus = - 2.5 m/s²

More Info

$\bigstar\ \; \green{\bf 1\ km/h=\dfrac{5}{18}\ m/s}$

Answer 2

(»•«)

Given:-

• Initial velocity (u) = 90 km/h = 25 m/s
• Final velocity (v) = 54 km/h = 15 m/s
• Time taken (t) = 4 s

{How can you change the velocities to km/h to m/s?

(»•«) → It's easy peasy buddy. Just multiply the speed value with (5/18)}

To Find: Acceleration (deceleration) of the particle.

We know,

a = (v - u)/t

where,

• a = Acceleration,
• v = Final velocity,
• u = Initial one &
• t = Time taken.

∴ a = [(15 - 25) m/s]/(4 s)

→ a = (- 10 m/s)/(4 s)

→ a = (- 2.5 m/s)/(s)

→ a = - 2.5 m/s²

☞ There was an acceleration of 2.5 m/s² opposing the direction of motion.