
A body is moving with a variable acceleration (a) along a straight line. The average acceleration of the body in time interval $ {t_1} $ to $ {t_2} $ is
Answer
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Hint: Total acceleration is what we are looking for in this question. Also during a straight-line motion, velocity and acceleration do not depend on the direction vector of the path as the direction can arbitrarily be assumed to be the +X axis.
Formulas used:
The average acceleration is the definite integral of instantaneous acceleration between the given limits of the time period divided by the total time divided by the total time of travel.
$ {a_{avg}} = \dfrac{{\int\limits_{{t_1}}^{_{{t_2}}} {adt} }}{{{t_{net}}}} $ where $ {a_{net}} $ is the net acceleration, a is the instantaneous acceleration and $ {t_1} $ to $ {t_2} $ is the time interval and $ {t_{net}} $ is the total time taken for travel.
Total time is nothing but the difference $ {t_1} $ and $ {t_2} $ .
$ {t_{net}} = {t_2} - {t_1} $
Complete step by step answer:
Here in this question, the acceleration is variable which means that it is changing with every interval of time. As such we need to calculate the sum of acceleration for all those intervals of time after breaking up the total travel duration into countless small parts. This can be done by integration. Also, the travel occurs in a limited time interval which is from $ {t_1} $ to $ {t_2} $ . As such the limits of integration will be $ {t_1} $ and $ {t_2} $ . As such the average acceleration is given by
$ {a_{avg}} = \dfrac{{\int\limits_{{t_1}}^{_{{t_2}}} {adt} }}{{{t_{net}}}} $
where $ {a_{net}} $ is the net acceleration, a is the instantaneous acceleration and $ {t_1} $ to $ {t_2} $ is the time interval and $ {t_{net}} $ is the total time taken for travel.
Time taken for this trip is the total time which is nothing but the $ {t_1} $ and $ {t_2} $ . It is given by
$ {t_{net}} = {t_2} - {t_1} $
As such the average acceleration will be
$ {a_{avg}} = \dfrac{{\int\limits_{{t_1}}^{_{{t_2}}} {adt} }}{{{t_2} - {t_1}}} $
Note: Here students make the error of calculating the average acceleration by just integrating the instantaneous acceleration with respect to time but that is not average acceleration rather it is a jerk.
Formulas used:
The average acceleration is the definite integral of instantaneous acceleration between the given limits of the time period divided by the total time divided by the total time of travel.
$ {a_{avg}} = \dfrac{{\int\limits_{{t_1}}^{_{{t_2}}} {adt} }}{{{t_{net}}}} $ where $ {a_{net}} $ is the net acceleration, a is the instantaneous acceleration and $ {t_1} $ to $ {t_2} $ is the time interval and $ {t_{net}} $ is the total time taken for travel.
Total time is nothing but the difference $ {t_1} $ and $ {t_2} $ .
$ {t_{net}} = {t_2} - {t_1} $
Complete step by step answer:
Here in this question, the acceleration is variable which means that it is changing with every interval of time. As such we need to calculate the sum of acceleration for all those intervals of time after breaking up the total travel duration into countless small parts. This can be done by integration. Also, the travel occurs in a limited time interval which is from $ {t_1} $ to $ {t_2} $ . As such the limits of integration will be $ {t_1} $ and $ {t_2} $ . As such the average acceleration is given by
$ {a_{avg}} = \dfrac{{\int\limits_{{t_1}}^{_{{t_2}}} {adt} }}{{{t_{net}}}} $
where $ {a_{net}} $ is the net acceleration, a is the instantaneous acceleration and $ {t_1} $ to $ {t_2} $ is the time interval and $ {t_{net}} $ is the total time taken for travel.
Time taken for this trip is the total time which is nothing but the $ {t_1} $ and $ {t_2} $ . It is given by
$ {t_{net}} = {t_2} - {t_1} $
As such the average acceleration will be
$ {a_{avg}} = \dfrac{{\int\limits_{{t_1}}^{_{{t_2}}} {adt} }}{{{t_2} - {t_1}}} $
Note: Here students make the error of calculating the average acceleration by just integrating the instantaneous acceleration with respect to time but that is not average acceleration rather it is a jerk.
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