a body is moving in a straight line along x-axis its distance from the origin is given by the equation x=at^2 bt^3 ,where x is in metre and T is in second find average speed of the body in the interval t is equal to zero and t is equal to 2 and its instantaneous speed at t is equal to 2 second.
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Solution-: A body is moving in a straight line along x-axis its distance from the origin is given by the equation x=at^2 bt^3.
Equation is x= t^2 - 4t + 6
x= t^2 - 4t +6
As we know that,
v= dx/dt
Therefore,
v = d (t^2 - 4t + 6)/dt = (2t - 4)
On substituting value of t as 0 and 3 sec we get
u= -4 m/s
v= 2 m/s
Now,
Average Speed= u+v/2
=>-4+2/2
= -2/2
= -1 m/s
Average Speed= -1 m/s.
So, speed is not in negative.
[So, Average Speed= 1 m/s. ]
Equation is x= t^2 - 4t + 6
x= t^2 - 4t +6
As we know that,
v= dx/dt
Therefore,
v = d (t^2 - 4t + 6)/dt = (2t - 4)
On substituting value of t as 0 and 3 sec we get
u= -4 m/s
v= 2 m/s
Now,
Average Speed= u+v/2
=>-4+2/2
= -2/2
= -1 m/s
Average Speed= -1 m/s.
So, speed is not in negative.
[So, Average Speed= 1 m/s. ]
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