A car moves along a straight line, whose equation of motion is given by S = 12 t + 3 t2 – 2 t3 , Where S is in metres and t is in seconds. The velocity of the car at start will be
Answers
Explanation:
S= 12t+3t²-2t³
S'= 12+6t-6t²
at starting t=0
= 12 +6×0-6×0
= 12 m/s
Given that, a car moves along a straight line and equation of motion of the car is given by:
s = 12t + 3t² - 2t³
Also given that, s is in meter and t is in seconds.
We have to find the velocity of the car at start.
Now,
Velocity = Displacement/Time
v = ds/dt
And given s is 12t + 3t² - 2t³
Substitute value of s in ds/dt
Velocity = d(12t + 3t² - 2t³)/dt
= 1*12 + 2*3t - 3*2t²
= 12 + 6t - 6t² .............(1st equation)
Velocity of the car at it's initial velocity will be 0 m/s. Also the time during starting/initial is 0 sec.
So, substitute value of t = 0 in (1st equation) to find the velocity of the car at start.
v = 12 + 6t - 6t²
v = 12 + 6(0) - 6(0)²
v = 12 + 0 - 6(0)
v = 12 + 0
v = 12
Therefore, the velocity of the car at the start will be 12 m/s.
Additional Information
Velocity = (Change in position)/(Change in time) = ds/dt
Acceleration = (Change in velocity)/(Change in time) = dv/dt
Velocity is a vector quantity as it has both magnitude and direction.
S.I. unit of velocity is m/s.
Dimensional formula of velocity = [L]/[T] = [L¹T-¹]