Math 2250 Lab 1 -Page 3 of 8 nsider the time-dependent velocity of the car shown
ID: 3194808 • Letter: M
Question
Math 2250 Lab 1 -Page 3 of 8 nsider the time-dependent velocity of the car shown in the figure below. (a) Describe the physical meaning of; (1) The graph of v(t), (2) The slope of the graph at a given point, '(t), (3) the area under the graph between two times (sds (b) Write down a piecewise function for v(t) using the graph. Write a sentence or two (c) Write down a piecewise function for r(t)-(s)ds using the graph. Write a (d) Find antiderivatives for your piecewise function on each of its pieces. Since there (e) Each time we choose these constants, the result will be a different function. How on how to tuse the graph to check the validity of your function. sentence or two on how you can use the graph to check the validity of this function. are four pieces, there will be four different constants of integration. can we choose these constants so that the resulting function is continuous provided that r(0) 0?Explanation / Answer
1:-
a) Physical meaning of
Graph v(t):- It depicts how the velocity of car varies with respect to time
Explaination:- For 0<t<1 the velocity of car keeps on rising with a specific rate as the rise is linear function of time. (ACCELERATION)
1<t<5 In this period the car keeps a steady speed or velocity throughout and i.e. 2km/min . (CONSTANT SPEED)
5<t<8 In this period velocity of car decreases with a constant rate from 2 km/min to 0.5 km/min.(DEACCELERATION)
8<t<10 Again the car maintains a constant speed throughout i.e. 0.5 km/min
Slope of graph v’(t):- It depicts the acceleration or deacceleration that the car undergoes.
Explaination:- For 0<t<1 change in v(t)/change in time
=(2-1)/(1-0) =+1km/min2 (ACCELERATION)
1<t<5 (2-2)/(5-1) = 0 km/min2 (CONSTANT SPEED)
5<t<8 (0.5- 2)/(8-5) = -0.5km/min2(Deacceleration)
8<t<10 (0.5-0.5)/10-8 =0km/min2 (constant speed)
Area under the graph between two points:- It will give the displacement covered till that time.
As v(t)dt = displacement covered within the time limits.
b) Piecewise function:-
For 0<t<1 the velocity of car keeps on rising with a specific rate as the rise is linear function of time. (ACCELERATION)
1<t<5 In this period the car keeps a steady speed or velocity throughout and i.e. 2km/min . (CONSTANT SPEED)
5<t<8 In this period velocity of car decreases with a constant rate from 2 km/min to 0.5 km/min.(DEACCELERATION)
8<t<10 Again the car maintains a constant speed throughout i.e. 0.5 km/min
For acceleration/deacceleration:-
0<t<1 change in v(t)/change in time
=(2-1)/(1-0) =+1km/min2 (ACCELERATION)
1<t<5 (2-2)/(5-1) = 0 km/min2 (CONSTANT SPEED)
5<t<8 (0.5- 2)/(8-5) = -0.5km/min2(Deacceleration)
8<t<10 (0.5-0.5)/10-8 =0km/min2 (constant speed)
c) how the acceration changes is already expresse. Again Iam expressing the same part:-
For acceleration/deacceleration:-
0<t<1 change in v(t)/change in time
=(2-1)/(1-0) =+1km/min2 (ACCELERATION)
1<t<5 (2-2)/(5-1) = 0 km/min2 (CONSTANT SPEED)
5<t<8 (0.5- 2)/(8-5) = -0.5km/min2(Deacceleration)
8<t<10 (0.5-0.5)/10-8 =0km/min2 (constant speed)
The slope of graph tells about acceleration/deacceleration or constant speed cases.