Question : If the velocity of a particle is v=At+Bt², where A and B are constants, then the distance travelled by it between 1s and 2s is
a) (3/2)A+(7/3)B
b) A/2 + B/3
c) (3/2)A+4B
d) 3A+7B.
Doubt by Trishna
Solution :
v=At+Bt² (Given)
We know
v=ds/dt
ds = vdt
Integrating both sides within proper limits
v=ds/dt
ds = vdt
Integrating both sides within proper limits
∫0sds = ∫12vdt
∫0sds = ∫12(At+Bt²)dt
∫0sds = ∫12(At+Bt²)dt
[s]0s = [At²/2+Bt³/3]12
s-0 = [A(2)²/2+B(2)³/3]-[A(1)²/2+B(1)³/3]
s-0 = [A(2)²/2+B(2)³/3]-[A(1)²/2+B(1)³/3]
s = [2A+8B/3]-[A/2+B/3]
s = [2A-A/2]+[8B/3-B/3]
s = [(4A-A)/2]+[(8B-B)/3]
s = [3A/2]+[7B/3]
s = (3/2)A+(7/3)B
s = [2A-A/2]+[8B/3-B/3]
s = [(4A-A)/2]+[(8B-B)/3]
s = [3A/2]+[7B/3]
s = (3/2)A+(7/3)B
Hence, a) (3/2)A+(7/3)B would be the correct option.