The velocity V ms^-1, of a moving body at time t seconds is given V = 5t^2 – 12t + 7
Calculate the acceleration when t = 2 seconds
To find the acceleration when t = 2 seconds, we need to find the derivative of the velocity function with respect to time.
Step 1: Find the first derivative of the velocity function
V = 5t^2 – 12t + 7
To find the derivative, we differentiate each term separately.
The derivative of 5t^2 with respect to t is 10t.
The derivative of -12t with respect to t is -12.
The derivative of 7 with respect to t is 0.
So, the derivative of V with respect to t is:
dV/dt = 10t - 12
Step 2: Substitute t = 2 into the derivative function
We substitute t = 2 into the derivative function we found in step 1.
dV/dt = 10(2) - 12
= 20 - 12
= 8
Therefore, the acceleration when t = 2 seconds is 8 m/s^2.
To calculate the acceleration when t = 2 seconds, we need to find the derivative of the velocity function with respect to time and then substitute t = 2 into the resulting expression.
Step 1: Find the derivative of the velocity function
The derivative of V with respect to t can be found by applying the power rule of differentiation. For each term, we multiply the coefficient by the power and then subtract 1 from the power.
V = 5t^2 – 12t + 7
dV/dt = d/dt(5t^2) – d/dt(12t) + d/dt(7)
dV/dt = 10t – 12 + 0
Simplifying, we have:
dV/dt = 10t – 12
Step 2: Substitute t = 2 into the derived expression
To find the acceleration at t = 2 seconds, we substitute t = 2 into the expression dV/dt = 10t – 12:
dV/dt = 10(2) – 12
dV/dt = 20 – 12
dV/dt = 8 m/s^2
Therefore, the acceleration when t = 2 seconds is 8 m/s^2.
Vo = 5*0^2 - 12*0 + 7 = 7 m/s. t = 0.
V = 5*2^2 - 12*2 * 7 = 3 m/s. t = 2 s.
a = (V-Vo)/t = (3-7)/2 = -2 m/s^2.