Max A. answered • 10/18/19

Professional Engineer with a Strong Tutoring/Academic Background

I'm going to rewrite the variables and equation a little clearer, to help understand this problem. Our equation is:

s = u*t + 1/2*a*t^{2}, where

s = distance traveled in meters (m)

t = time in seconds (s)

u = initial velocity in meters per second (m/s)

a = acceleration in meters per second squared (m/s^{2})

One more thing to note is that our equation for distance traveled is a function of time, s(t).

(a) I wont be able to draw a graph on this platform, but one thing we can do is determine a few points on the curve. Then you can connect these points to make a smooth curve.

u = 1 m/s (I am assuming a value of 1, since it isn't actually provided in your question)

a = 5 m/s^{2}

s(t) = (1)*t + 1/2*(5)*t^{2}

s(t) = t + 2.5*t^{2}

s(0) = 0 + 2.5*(0)^{2} = 0, corresponds to (0,0)

s(1) = 1 + 2.5*(1)^{2} = 3.5, corresponds to (1,3.5)

s(2) = 2 + 2.5*(2)^{2} = 12, corresponds to (2,12)

etc.......

s(10) = 10 + 2.5*(10)^{2} = 260, corresponds to (10,260)

(b) The gradient of a curve at any point is given by the slope of the tangent line at that point. Basically, draw a tangent line to the curve at the time(s) in question (at t=2 and t=6) and calculate the slopes of these tangent lines (drawing the graph to scale will be critical if you want to do this by hand).

(c) i)

s(t) = t + 2.5*t^{2}

v(t) = s'(t) = ds/dt = 1 + 5t

ii) a(t) = v'(t) = s''(t) = d^{2}s/dt^{2} = 5

(d) v(t) = 1 + 5t

v(2) = 1 + 5*(2) = 11

v(6) = 1 + 5*(6) = 31

(e) In part (b), we are drawing tangent lines to the curve and calculating the slopes of those tangent lines. Does that sound familiar to another concept we learn in calculus? What are we actually doing when we take a derivative of a function/curve? I will let you draw your own conclusions.

Baneen A.

Wow, the tutor that helps me with my Questions only rights 5% of what you right11d