The displacement (in meters) of an object moving in a straight line is given by

Question

The displacement (in meters) of an object moving in a straight line is given by S(t) = 9 + 3t - 2t^2, where t is measured in seconds. Find the average velocity over the following lime periods: Use the result obtained Iron (a) to estimate the instantaneous velocity when t = 1 sec. Then find the exact value of the instantaneous velocity at t = 1 sec. What is the acceleration of the object at t = 1 sec.? Let f(x) = {x^2 + x + 1, if x 2; Sketch the graph of y = f(x). Show that f(x) is continuous at x = 2; Determine whether f is differentiable at x = 2. Explain why or why not.

Explanation / Answer

S(t) = 9 + 3t - 2t2

a) i) [1 , 1.03]

Average velocity = [ S(1.03) - S(1) ] / [ 1.03 - 1]

==> Average velocity = [ 9 + 3(1.03) - 2(1.03)2 - (9 + 3(1) - 2(1)2) ] / 0.03

==> Average velocity = [ 9.9682 - (10) ] / 0.03 = -1.06

==> Average velocity over [1 , 1.03] is -1.06 m/s

ii) [1 , 1.0003]

Average velocity = [ S(1.0003) - S(1) ] / [ 1.0003 - 1]

==> Average velocity = [ 9 + 3(1.0003) - 2(1.0003)2 - (9 + 3(1) - 2(1)2) ] / 0.0003

==> Average velocity = [ 9.9997 - (10) ] / 0.0003 = -1.06

==> Average velocity over [1 , 1.0003] is -1.06 m/s

iii) [1 , 1.000003]

Average velocity = [ S(1.000003) - S(1) ] / [ 1.000003 - 1]

==> Average velocity = [ 9 + 3(1.000003) - 2(1.000003)2 - (9 + 3(1) - 2(1)2) ] / 0.000003

==> Average velocity = [ 9.999997 - (10) ] / 0.000003 = -1.000006

==> Average velocity over [1 , 1.000003] is -1.000006 m/s

b) Instantaneous velocity = -1 m/s . since the average velocity value is tending towards -1

c) S(t) = 9 - 3t - 2t2

V(t) = d S(t) / dt = 0 - 3(1) - 2(2) t2-1 = -3 - 4t             since d/dx xn = n xn-1

Acceleration a(t) = d V(t) / dt = 0 - 4(1) = -4

Hence acceleration at t = 1 sec is -4 m/s2

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