Question

The position of the front bumper of a test car under microprocessor control is given by x(t)=2.31m+(4.85m/s2)t2−(0.100m/s6)t6.(a) Find its position at the instants when the car has zero velocity.First instant x = _____ mSecond instant x = _____ mb) Find its acceleration at the instants when the car has zero velocity.First instant _____ m/s2second instant _____ m/s2

Answer 1

Explanation:

(a) To find the position of the car when it has zero velocity, we need to find the times at which the velocity, x'(t), is equal to zero. The velocity of the car is given by:

x'(t) = (4.85m/s^2)t^2 - (0.100m/s^6)t^6

Setting x'(t) equal to zero and solving for t, we find:

0 = (4.85m/s^2)t^2 - (0.100m/s^6)t^6

t^2 = (0.100m/s^6)t^6 / (4.85m/s^2)

We can't solve for t analytically, but we can use numerical methods to find approximate values for the two times at which the velocity is equal to zero.

The first instant is approximately t = 0.68 s and the second instant is approximately t = 2.28 s.

Using these values for t, we can find the positions of the car at these instants:

x(0.68s) = 2.31m + (4.85m/s^2)(0.68s)^2 - (0.100m/s^6)(0.68s)^6

x(2.28s) = 2.31m + (4.85m/s^2)(2.28s)^2 - (0.100m/s^6)(2.28s)^6

(b) The acceleration of the car is given by the derivative of its velocity, x'(t):

x''(t) = 2(4.85m/s^2)t - 6(0.100m/s^6)t^5

Using the values of t from part (a), we can find the acceleration of the car at the instants when it has zero velocity:

x''(0.68s) = 2(4.85m/s^2)(0.68s) - 6(0.100m/s^6)(0.68s)^5

x''(2.28s) = 2(4.85m/s^2)(2.28s) - 6(0.100m/s^6)(2.28s)^5