Question : The position of an object moving along x-axis is given by x=a+bt² where a= 8.5m, b = 2.5 m/s² and t is measured in seconds. What is its velocity at t=0s and t=2.0s. What is the average velocity between t=2.0s and t=4.0s?
Doubt by Saniya
Solution :
x = a + bt²
a=8.5m,
b = 2.5 m/s²
b = 2.5 m/s²
Velocity at t=0 sec
V=dx/dt
V=d(a+bt²)/dt
V=0+2bt
V=2bt
Velocity at t = 0sec
V=2b(0)
V=0 m/s
V=dx/dt
V=d(a+bt²)/dt
V=0+2bt
V=2bt
Velocity at t = 0sec
V=2b(0)
V=0 m/s
Velocity at t = 2sec
V=2b(2)
V=4b
V=4(2.5)
V=10 m/s
V=2b(2)
V=4b
V=4(2.5)
V=10 m/s
Average velocity between 2sec and 4sec
x=a+bt²
Distance at t = 2sec
x1=a+b(2)²
x1=a+4b
Distance at t = 4sec
x2=a+b(4)²
X2=a+16b
Average velocity
Vav=(x2-x1)/(t2-t1)
Vav=[(a+16b)-(a+4b)]/(4-2)
Vav=[a+16b-a-4b]/2
Vav=12b/2
Vav=6b
Vav=6×2.5
Vav=15 m/s
Distance at t = 2sec
x1=a+b(2)²
x1=a+4b
Distance at t = 4sec
x2=a+b(4)²
X2=a+16b
Average velocity
Vav=(x2-x1)/(t2-t1)
Vav=[(a+16b)-(a+4b)]/(4-2)
Vav=[a+16b-a-4b]/2
Vav=12b/2
Vav=6b
Vav=6×2.5
Vav=15 m/s