272.0
If a body travels with an acceleration a, for time t, and acceleration a, for time tz, t, and t,
being successive time intervals, then the average acceleration of the body is
Answers
Answer:
Equations of motion for a uniform accelerated body changing its velocity u to v in time t, with an acceleration a and covering a displacement s in the time.
v = u + at
s = ut + at²/2
v² = u² + 2as
Solution
Let the initial velocity of the body be u,
The body travels with acceleration a₁ for time t₁
Final Velocity for this time period is,
v = u + a₁t₁
Now, The body accelerates with a₂ for time t₂
The velocity after this time interval is given,
w = v + a₂t₂
w = u + a₁t₁ + a₂t₂
Average acceleration :
It is the change in velocity over the time interval.
If velocity changes from a to b over the time interval t,
\frac{a - b}{t} = avg \: acceleration
t
a−b
=avgacceleration
Now Average acceleration is the change in velocity over the entire time interval.
Time t = t₁+ t₂
Initial velocity = u
Final Velocity = w
So, Average acceleration is,
\begin{gathered}= \frac{w - u}{t} \\ \\ = \dfrac{u + a_1t_1+ a_2t_2 - u}{t} \\ \\ = \dfrac{ a_1t_1+ a_2t_2 }{t} \\ \\ = \dfrac{ a_1t_1+ a_2t_2 }{t_1+ t_2}\end{gathered}
=
t
w−u
=
t
u+a
1
t
1
+a
2
t
2
−u
=
t
a
1
t
1
+a
2
t
2
=
t
1
+t
2
a
1
t
1
+a
2
t
2
Therefore, The average acceleration of the body is
\dfrac{ a_1t_1+ a_2t_2 }{t_1+ t_2} \:
t
1
+t
2
a
1
t
1
+a
2
t
2