Question

A car going at 24 ft/s decelerates at a constant 4ft/s^2 to come to a stop.

a) Fill in the following table showing the velocity of the car every second. Fill in a zero velocity for any time after that at which the car comes to a rest.

t =	0	1	2	3	4	5	6	7
v =	24	20	16	12	8	4	0	0

b) When does the car come to a stop?

at t = ___6___

Using your table, find left and right sums which estimate the total distance traveled before the car comes to rest. Which is an overestimate, and which is an underestimate?

The __right sum__ is an underestimate, and is = __60__

The __left sum__ is an overestimate, and is = __84__

c) On a sheet of paper, sketch a graph of velocity against time. On the graph, show an area representing the distance traveled before the car comes to rest. Use the graph to calculate this distance.

distance = ___?___

d) Now find a formula for the velocity of the car as a function of time:

velocity = ___?___

and then find the total distance traveled by antidifferentiation.

distance = ___?___

Answer 1

(c) the graph will looks like

From the velocity-time graph, the distance will be the area under the curve. Therefore, the area will be equal to the area of the shaded triangle (OAB) in the above figure.

(d) The formula for velocity with time can be calculated from the equation of the straight line AB. The general equation for straight line is y = mx + c, where m is the slope and c is the y intercept. Here, c is equal to 24 ft/s (at x = 0, the value of y). The slope is equal to

Here, the v is plotted in y-axes and time t is plotted in x-axes. Therefore, the equation for the straight line will be

If we apply the antidifferentiation i.e. integration method, you can obtain

As the car comes to rest after 6 s, the distance can be calculated by putting the value of t as 6 s. Therefore, form the above equation, you can obtain the distance (S)