the position of a particle moving on X axis is given by x=At^3+Bt^2+Ct+D.the numwrical values of A,B,C,D are 1,4,-2,5 respectively and SI units are used.
find a)the dimensons of A,B,C,D.
b)the velocity of the particle at t=4s
c)the acceleration of the particle at t=4s
d)the average velocity during the interval t=0 to t=4 s
e)the avreage acceleration during the interval t=0 to t=4s
Answers
Answer:
x=At³ + Bt² + Ct + D
A = 1 , B = 4 , C = -2 , D = 5
some basic derivatives needed for this answer have been given in the following attachment
If any doubt please check with it
a) the dimensions of A,B,C,D
x=At³ + Bt² + Ct + D
since x is a length
At³ = L = L¹t⁻³
Bt² = L = L¹t⁻²
Ct = L =L¹t¹
D = L = L¹
b) the velocity of the particle at t=4s
x=At³ + Bt² + Ct + D
v = dx / dt
v = d At³ + Bt² + Ct + D / dt
v = 3At² + 2Bt + C
at t = 4s v = 3A×4² + 2B×4 + C
V = 48A + 8B + C
V = 48 × 1 + 8 × 4 - 2
V = 48 + 32 - 2
V = 78ms⁻¹
c)the acceleration of the particle at t=4s
v = 3At² + 2Bt + C
a = dv / dt
a = d 3At² + 2Bt + C / dt
a = 3A × 2t + 2B + 0
a = 6At + 2B
at t = 4s a = 6A×4 + 2B
a = 24A + 2B
a = 24 + 2×4
a = 32ms⁻²
d) the average velocity during the interval t=0 to t=4 s
x=At³ + Bt² + Ct + D
Vav = total distance / total time
total distance
= displacement from t=0s to t=4s
= [A×4³+ B×4²+ C×4 +D] - [A × 0 + B × 0 + C × 0 + D]
= 64A + 16B + 4C + D - D
= 64 + 16×4 - 8
= 64 + 64 - 8
= 120m
Vav = 120 / 4
= 30ms⁻¹
e) the average acceleration during the interval t=0 to t=4s
v = 3At² + 2Bt + C
Average acceleration= total change in velocity / time
total change in velocity
= velocity from t=0s to t=4s
= [3A×4² + 2B×4+ C] - [3A×0 + 2B×0 + C]
= 48A + 8B + C - C
= 48A + 8B
= 48 + 32
= 80ms⁻¹
Average acceleration = 80 / 4
= 20ms⁻²
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