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

[albodibran]

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

 

[berkeman]

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...

$$v(t) = v_0 + at$$