x=5t^3-6t^2-7t meter, what is the displacement from t1=1s to t2=3s. What is the instantaneous velocityand instantaneous acceleration?

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The position x in terms of time t is given as x = 5t^3 - 6t^2 - 7t

Displacement from t = 1 to t = 3 is equal to:

5*3^3 - 6*3^2 - 7*3 - (5*1^3 - 6*1^2 - 7*1)

=> 60 - (-8)

=> 68 meters

The instantaneous velocity is the derivative of x with respect to time. dx/dt = 15*t^2 -12t - 7 m/sThe instantenous acceleration is the derivative of velocity with respect to time. Here it is equal to 30t - 12 m/s^2

The displacement from t=1 to t=3 is the given function f(t)'s

f(3)-f(1)=60-(-8)=68 meters

the instantaneous velocity is the derivative of x with respect to t

in math notation

dx/dt

dx/dt the original equation

using the power rule

v(t)= 15t^2-12t-7 with t measured in seconds and velocity measured in meters/second

the instantaneous acceleration is the derivative of the velocity(the slope of the velocity graph) with respect to time

dx/dt

dx/dt the velocity equation

using the power rule again

the constant -7 becomes 0

a(t)=30t-12 measured in m/s^2

The final answer to this question

**The displacement from t=1 to t=3 is 68m, the instantaneous velocity is v(t)=15t^2-12t-7**

**the instantaneous acceleration function is a(t)= 30t-12**

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