a car, starting from rest , began moving in straight line and its velocity is given by v= (8t2+6t) m/s then its displacement at t=9 equals
To find the displacement at a specific time, we need to integrate the velocity function with respect to time.
∫v(t) dt = ∫(8t^2 + 6t) dt = 8(t^3/3) + 6(t^2/2) + C
where C is the constant of integration.
Since the car started from rest, its initial position is 0. Therefore, the displacement at t=9 can be found by evaluating the above expression at t=9 and subtracting the initial position:
displacement at t=9 = 8(9^3/3) + 6(9^2/2) - 0 = 2052 meters
Therefore, the displacement of the car at t=9 is 2052 meters.
1944 + 243 = 2187
That is correct.
2187 is not 2052 meters
You are correct. I apologize for the mistake in my previous response. I will recalculate the displacement at t=9:
∫v(t) dt = ∫(8t^2 + 6t) dt = 8(t^3/3) + 6(t^2/2) + C
At t=0, the car starts from rest, therefore, initial velocity = u = 0
Therefore, the constant of integration is C=0.
Putting this value in above equation we get,
Displacement (d) = 8(9^3/3) + 6(9^2/2) + 0 = 2187 meters
Therefore, the displacement of the car at t=9 is 2187 meters. I apologize for any confusion caused.