Calculating distance per 1-sec interval for plot graph d vs t

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SUMMARY

The discussion focuses on calculating the distance traveled by a car accelerating from an initial velocity of 14 m/s to a final velocity of 24 m/s with a constant acceleration of 2.0 m/s². The time taken to reach this velocity is determined to be 5 seconds using the formula t = (v - v0) / a. The total distance covered during this acceleration phase is calculated to be 95 meters using the equation d = v0(t) + 1/2(at²). Additionally, the participants discuss computing distance at 1-second intervals, concluding that the distance per second averages 19 m, leading to a distance-time graph.

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albodibran
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2 travelining in a straight line initial velocity is 14m/s, at a constant acceleration of 2.0m/s^2 to a velocity of 24 m/s

a) how much time to reach velocity of 24m/s

t= v-v0/a or 24-14/2= 5 seconds

b) distance covered by car in this process?

d=v0(t) + 1/2(at^2) or (14)(5)+1/2(2)(5^2)= 70m+25m= 95m

c) compute values of the distance traveled at 1-sec intercals and carefully draw a graph of distance plotted against time for this motion?
c is the only one throwing me off

the problem doesn't say it is constant but this is the approach I took to form a graph plot

so then distance per second would be 19 m?
by taking distance/time 95m/5s
so at each second interval it would be like this t graph?

t-0 d-0
t-1 d-19
t-2 d- 38

so on and so forth?
I mean it doesn't say its constant acceleration so idk if what I did was right for c
 
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albodibran said:
2 travelining in a straight line initial velocity is 14m/s, at a constant acceleration of 2.0m/s^2 to a velocity of 24 m/s

a) how much time to reach velocity of 24m/s

t= v-v0/a or 24-14/2= 5 seconds

b) distance covered by car in this process?

d=v0(t) + 1/2(at^2) or (14)(5)+1/2(2)(5^2)= 70m+25m= 95m

c) compute values of the distance traveled at 1-sec intercals and carefully draw a graph of distance plotted against time for this motion?
c is the only one throwing me off

the problem doesn't say it is constant but this is the approach I took to form a graph plot

so then distance per second would be 19 m?
by taking distance/time 95m/5s
so at each second interval it would be like this t graph?

t-0 d-0
t-1 d-19
t-2 d- 38

so on and so forth?
I mean it doesn't say its constant acceleration so idk if what I did was right for c

On a), I don't think you used the equation correctly. Start with the equation for velocity and carefully apply your numbers...

v(t) = v_0 + at
 

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