A bus increases its velocity from 36km/hr to 54km/hr in 5 seconds.Calculate the acceleration
Answers
Given :
- Initial velocity of the bus = 36 km/hr
- Final velocity of the bus = 54 km/hr
- Time = 5 sec
To Find :
- Acceleration of the bus
Solution :
The relation between km/hr and m/s is ,
Now ,
Initial velocity = 36 km/hr = 36 × 0.278 = 10 m/s
Final velocity = 54 km/hr = 54 × 0.278 = 15 m/s
The relation between acceleration , Initial velocity , Final velocity and time is given by ,
Where ,
- v is final velocity
- u is initial velocity
- t is time
We have ,
- v = 54 km/hr =15m/s
- u = 36 km/hr = 10 m/s
- t = 5 s
∴ The Acceleration of the bus when the velocity changes from 36 km/hr to 54 km/hr in 5 sec is 1 m/s²
Answer:
Given :
Initial velocity of the bus = 36 km/hr
Final velocity of the bus = 54 km/hr
Time = 5 sec
To Find :
Acceleration of the bus
Solution :
The relation between km/hr and m/s is ,
\boxed{ \rm{km. {hr}^{ - 1} = 0.278 \: m {s}^{ - 1} }}
km.hr
−1
=0.278ms
−1
Now ,
Initial velocity = 36 km/hr = 36 × 0.278 = 10 m/s
Final velocity = 54 km/hr = 54 × 0.278 = 15 m/s
The relation between acceleration , Initial velocity , Final velocity and time is given by ,
\dag \boxed {\rm{a = \frac{v - u}{t} }}†
a=
t
v−u
Where ,
v is final velocity
u is initial velocity
t is time
We have ,
v = 54 km/hr =15m/s
u = 36 km/hr = 10 m/s
t = 5 s
\begin{gathered} : \implies \rm \: a = \frac{15 \: m {s}^{ - 1} - 10 \: ms {}^{ - 1} }{5 \: s} \\ \\ : \implies \rm \: a = \frac{5 \: m {s}^{ - 1} }{5 \: s} \\ \\ : \implies \rm \: a = 1 \: m {s}^{ - 2} \: \bigstar\end{gathered}
:⟹a=
5s
15ms
−1
−10ms
−1
:⟹a=
5s
5ms
−1
:⟹a=1ms
−2
★
∴ The Acceleration of the bus when the velocity changes from 36 km/hr to 54 km/hr in 5 sec is 1 m/s²