Question

a bus decreases its speed from 72 km/hr to 54 km/hr in 10 seconds. calculate the acceleration of the bus.

Answer 1

u=72 km/hr=72*5/18 m/s=20 m/s
v=54 km/hr=54*5/18 m/s=15 m/s
a=d/dt(v)
=>a=15-20/10
=>a=-5/10
=>a=-0.5 m/s^2
therefore,negative acceleration i.e.,deceleration of the bus is 0.5 m/s^2

Answer 2

ACCELERATION IS -0.5 m/s².

Given:

• Initial speed = 72 km/hr
• Final speed = 54 km/hr
• Time = 10 sec

To find:

• Acceleration of bus ?

Calculation:

First let's convert the unit of velocity from km/hr to m/s.

$u = 72 \: kmph = 72 \times \dfrac{5}{18 } = 20 \: m/s$

$v= 54 \: kmph = 54 \times \dfrac{5}{18 } = 15 \: m/s$

Now, applying equations of kinematics:

$a = \dfrac{v - u}{t}$

$\implies a = \dfrac{15- 20}{10}$

$\implies a = \dfrac{ - 5}{10}$

$\implies a = - 0.5 \: m/ {s}^{2}$

So, acceleration is - 0.5 m/s²