a) 5 sec
b) 4 sec
b) 4 sec
c) 1/5 sec
d) 1/10 sec
Doubt by Sarthak
Solution :
v(t)=25t2-10t+5
differentiating both sides w.r.t. time.
dv/dt = 50t-10 — (1)
dv/dt=0
0=50t-10
v(t)=25t2-10t+5
differentiating both sides w.r.t. time.
dv/dt = 50t-10 — (1)
dv/dt=0
0=50t-10
10=50t
t=10/50
t = 1/5 sec
t=10/50
t = 1/5 sec
Again differentiating equation (1) w.r.t. time
d²v/dt² = 50
∴ d²v/dt² >0
Hence the function has minimum velocity at time t = 1/5 sec and the minimum velocity is given by
v(1/5) = 25(1/5)2-10(1/5)+5
v(1/5) = 25/25-2+5
v(1/5) = 1-2+5
v(1/5) = 4 m/s
d²v/dt² = 50
∴ d²v/dt² >0
Hence the function has minimum velocity at time t = 1/5 sec and the minimum velocity is given by
v(1/5) = 25(1/5)2-10(1/5)+5
v(1/5) = 25/25-2+5
v(1/5) = 1-2+5
v(1/5) = 4 m/s
It simply means that the function has no absolute maximum value. So we need to find the value of velocity at time t=5 sec, t=4 sec and t=1/10 sec.
So,
v(t)=25t2-10t+5
v(5)=25(5)2-10(5)+5
= (25×25)-50+5
= 625-45
= 580 m/s
So,
v(t)=25t2-10t+5
v(5)=25(5)2-10(5)+5
= (25×25)-50+5
= 625-45
= 580 m/s
v(4)=25(4)2-10(4)+5
= (25×16)-40+5
= 400-35
= 365 m/s
v(1/5) = 4 m/s
v(1/10)=25(1/10)2-10(1/10)+5
= (25/100)-1+5
= 1/4+4
= 0.25+4
= 4.25 m/s
Clearly the velocity of the particle is maximum at time t=5 sec.
Hence a) 5 sec is the correct option.
= (25/100)-1+5
= 1/4+4
= 0.25+4
= 4.25 m/s
Clearly the velocity of the particle is maximum at time t=5 sec.
Hence a) 5 sec is the correct option.
PS : Here is the graph of the velocity function for your reference.
