Answer to Question #99175 in Calculus for M

Question #99175
**THERE MUST BE A GRAPHICAL METHOD FOR CALCULATING THE GRADIENT**

Question:
The 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 / 2 at^2

A) plot a graph showing a graphical method for calculating gradient of distance (s) vs time (t) for the first 10 seconds 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 your result from part c to calculate the velocity at t= 2s and t= 6s?

E) compare results from parts b and d?
1
Expert's answer
2019-11-27T09:48:04-0500

A)"S=10t+5t^2\/2"

"S=(5\/2)[(t+2)^2-10]"

At "t=10" ,

From the graph,"x- intercept=4.41"

"y-intercept=-265"

"dS\/dt=265\/4.41=60.1 m\/s"

B)At "t=2s"

From the graph,

"x- intercept=1.25"

"y- intercept=-25"

"dS\/dt=20m\/s"

At "t=4s"

"x- intercept=1.83"

"y- intercept=-55"

"dS\/dt=55\/1.83=30.054m\/s"

C) "S=ut+(1\/2)at^2"

"dS\/dt=u+at"

"v=u+at"

"dv\/dt=a"

"acceleration=a"

D)"v=u+at"

At "t=2s"

"v=10+5\u00d72=20m\/s"

At "t=4s"

"v=10+5\u00d74=30m\/s"

E)The answers obtained in Part B and Part D are approximately equal.

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