Answer to Question #97525 in Calculus for Matt

Question

Equation for a distance, s(m), travelled in time t(s) by an object starting with an initial velocity u(ms^-1) and uniform acceleration a(ms^-2) is:

s=ut+1/2at^2

a) plot a graph of distance (s) vs time (t) for the first 10s of motion if u = 10ms^-1 and a = 5ms^-2?

b) determine the gradient of the graph at t=2s and t=6s?

c) differentiate the equation to find the functions for
I) velocity (v=ds/dt)
Ii) acceleration (a= dv/dt = d^2 s/dt^2

d) use result from part c to calculate the velocity at t=2s and t=6s?

e) compare your results for part b and d?

Answer 1

a) Putting the value of a and u in the equation s=ut +1/2at^2, we get

S=10t + 5/2 t^2

Now plotting this graph,we get

b) Gradient of any graph can be calculated by using derivatives

So, de/dt = 10 + 5t

At t= 2s, gradient= 10+ 5*2 = 20

At t=6s, gradient= 10+ 5*6 = 40

c)

I) Velocity, V= ds/dt = 10+5t

ii) Acceleration, a = dv/dt = 5

d) we have V = 10+5t

At, t = 2s, V = 10+5*2 = 20 m/s

At, t= 6s, V = 10+5*6 = 40 m/s

e) We can see that the solution from part b and part d is same. So,it can

be concluded that the gradient of the Distance vs Time graph is velocity.