Question

linear momentum of a particle moving along a straight line as a function of time is given as p=p0e^-alphat^3 where p0 and alpha are constants. Time is measured with a stop watch of least count 10^-2 s and value of alpha is 1s^-3. The percentage error in the measurement of p at t=1s is​

Answer 1

Given info : linear momentum of a particle moving along a straight line as a function of time is given as $P=P_0e^{-\alpha t^3}$ where P₀ and alpha are constants. Time is measured with a stop watch of least count 10¯² s and value of alpha is 1s¯³

To find : The percentage error in the measurement of P at t = 1s is ....

solution : here P = P₀e^(-αt³)

taking log base e both sides,

lnP = lnP₀ - αt³

differentiating both sides we get,

dP/P = 0 -3αt² dt/t

⇒% error in P = 3αt² × % error in t

here % error in t = ∆t/t × 100 = 10¯²/1 × 100 = 1 %

t = 1 s , α = 1 /s³

so, % error in P = 3 × 1 × 1² × 1% = 3 %

Therefore the percentage error in the measurement of P at t = 1s is 3 %.

also read similar questions : The position s of a particle moving along a straight line at time t is given by s = at3+ bt5. Find its

velocity at t=1s....

https://brainly.in/question/19132321

the linear momentum of a particle as a function of time is given by P is equals to a + b t where A and B are positive co...

https://brainly.in/question/15492802