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New Basic Physics
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Q. An object moves along the x−axis. Its x−coordinates is given as a function of time as x=7t−3t 2 m where x is in meters and t is in seconds. Its average speed over the interval t=0 to t=4 s is?
X=7t−3t 2 ⇒v= dt dx=7−6t Let us find the time when the velocity becomes zero Putting v=0⇒7−6t=0 ⇒t=7/6 s x (t−7/6 s)=7× 6 7−3× 6 2 7 2= 12 49m At t=0→x=0 At t=4 s→x=7×4−3(4) 2 =−20 m Distance travelled =s 12 49+ 12 49+20= 6 169ms −1
Q. The relation between time t and displacement x is t = αx^2+βx, where α and β are constants. The retardation is I have successfully differentiated till velocity but from differentiating acceleration why we are using d(x^m + N)^n/dx identity but why not d(x+N) ^n/dx, since the x in velocity is linear.?
you will get velocity in terms of x , then to find accleration you can use chain rule formula
a = v*dv/dx
Q. what is meant by potential gradient?
potential gradient dV / dx is difference in potential between two points, or per unit length
Q. In acceleration how to use a= v*dv/ds and In which kind of example? And what is the difference between a = v*dv/ds and a = dv/dt??
accleration is the rate of change in velocity
a = dv / dt
now using chain rule
a = dv / ds * ds /dt [ where ds is the change in displacement and ds / dt = velocity ]
so a = dv / ds * v
so when velocity is given as function of s , then we can use this formula
Q. An object moves along x axis. It's X coordinates is given as a function of time as x = 7t - 3tsquare.Find the average speed from t = 0 to t = 4s?
vavg = ∆x/ ∆t = x2 − x1 / t2 − t1
where x1(t = 0) = 0;
x2(t = 4) = 7(4) − 3(4)2 = −20 m
So; vavg = −20 − 0 / 4 − 0 = −5.00 m/s