Linear model slope gives NA when reducing original data

I have a df with a response variable that changes with time. When plotting the data, there is a plateau of which I want to get rid of, since I need the steepest slope of the regression:

time=c("1899-12-31 09:11:37 UTC", "1899-12-31 09:12:34 UTC", "1899-12-31 09:13:04 UTC", "1899-12-31 09:13:34 UTC", "1899-12-31 09:14:04 UTC","1899-12-31 09:14:34 UTC", "1899-12-31 09:15:04 UTC", "1899-12-31 09:15:34 UTC", "1899-12-31 09:16:04 UTC", "1899-12-31 09:16:34 UTC","1899-12-31 09:17:04 UTC", "1899-12-31 09:17:34 UTC", "1899-12-31 09:18:04 UTC", "1899-12-31 09:18:34 UTC", "1899-12-31 09:19:04 UTC","1899-12-31 09:19:34 UTC", "1899-12-31 09:20:04 UTC", "1899-12-31 09:20:34 UTC" ,"1899-12-31 09:21:04 UTC" ,"1899-12-31 09:21:34 UTC","1899-12-31 09:22:04 UTC", "1899-12-31 09:22:34 UTC", "1899-12-31 09:23:04 UTC" ,"1899-12-31 09:23:34 UTC", "1899-12-31 09:24:04 UTC","1899-12-31 09:24:34 UTC", "1899-12-31 09:25:04 UTC" ,"1899-12-31 09:25:34 UTC" ,"1899-12-31 09:26:04 UTC" ,"1899-12-31 09:26:34 UTC","1899-12-31 09:27:04 UTC", "1899-12-31 09:27:34 UTC" ,"1899-12-31 09:28:04 UTC", "1899-12-31 09:28:34 UTC" ,"1899-12-31 09:29:04 UTC","1899-12-31 09:29:34 UTC" ,"1899-12-31 09:30:04 UTC" ,"1899-12-31 09:30:34 UTC" ,"1899-12-31 09:31:04 UTC", "1899-12-31 09:31:34 UTC","1899-12-31 09:32:04 UTC", "1899-12-31 09:32:34 UTC" ,"1899-12-31 09:33:04 UTC", "1899-12-31 09:33:34 UTC", "1899-12-31 09:34:04 UTC","1899-12-31 09:34:34 UTC" ,"1899-12-31 09:35:04 UTC" ,"1899-12-31 09:35:35 UTC", "1899-12-31 09:36:04 UTC", "1899-12-31 09:36:35 UTC","1899-12-31 09:37:04 UTC", "1899-12-31 09:37:35 UTC" ,"1899-12-31 09:38:04 UTC" ,"1899-12-31 09:38:35 UTC", "1899-12-31 09:39:04 UTC","1899-12-31 09:39:35 UTC" ,"1899-12-31 09:40:04 UTC" ,"1899-12-31 09:40:35 UTC" ,"1899-12-31 09:41:04 UTC" ,"1899-12-31 09:41:35 UTC") var1=c( 7.44, 7.41, 7.40, 7.39 ,7.38, 7.36, 7.36,7.35 ,7.34, 7.34, 7.33, 7.33 ,7.32 ,7.32 ,7.32 ,7.31 ,7.31 ,7.31, 7.31 ,7.31 ,7.30 ,7.30, 7.30 ,7.30 ,7.30 ,7.30 ,7.30,         7.30 ,7.30, 7.30, 7.30, 7.30, 7.30, 7.30, 7.30, 7.30, 7.30 ,7.30 ,7.30, 7.30, 7.30, 7.30, 7.30, 7.30, 7.30, 7.30, 7.30 ,7.30, 7.30, 7.30 ,7.30,7.30 ,7.30, 7.30,         7.30 ,7.30 ,7.30, 7.30 ,7.30 ,7.30) df<-data.frame(time,var1) df$time<-as.POSIXct(df$time) df                  time var11  1899-12-31 09:11:37 7.442  1899-12-31 09:12:34 7.413  1899-12-31 09:13:04 7.404  1899-12-31 09:13:34 7.395  1899-12-31 09:14:04 7.386  1899-12-31 09:14:34 7.367  1899-12-31 09:15:04 7.368  1899-12-31 09:15:34 7.359  1899-12-31 09:16:04 7.3410 1899-12-31 09:16:34 7.3411 1899-12-31 09:17:04 7.3312 1899-12-31 09:17:34 7.3313 1899-12-31 09:18:04 7.3214 1899-12-31 09:18:34 7.3215 1899-12-31 09:19:04 7.3216 1899-12-31 09:19:34 7.3117 1899-12-31 09:20:04 7.3118 1899-12-31 09:20:34 7.3119 1899-12-31 09:21:04 7.3120 1899-12-31 09:21:34 7.3121 1899-12-31 09:22:04 7.3022 1899-12-31 09:22:34 7.3023 1899-12-31 09:23:04 7.3024 1899-12-31 09:23:34 7.3025 1899-12-31 09:24:04 7.3026 1899-12-31 09:24:34 7.3027 1899-12-31 09:25:04 7.3028 1899-12-31 09:25:34 7.3029 1899-12-31 09:26:04 7.3030 1899-12-31 09:26:34 7.3031 1899-12-31 09:27:04 7.3032 1899-12-31 09:27:34 7.3033 1899-12-31 09:28:04 7.3034 1899-12-31 09:28:34 7.3035 1899-12-31 09:29:04 7.3036 1899-12-31 09:29:34 7.3037 1899-12-31 09:30:04 7.3038 1899-12-31 09:30:34 7.3039 1899-12-31 09:31:04 7.3040 1899-12-31 09:31:34 7.3041 1899-12-31 09:32:04 7.3042 1899-12-31 09:32:34 7.3043 1899-12-31 09:33:04 7.3044 1899-12-31 09:33:34 7.3045 1899-12-31 09:34:04 7.3046 1899-12-31 09:34:34 7.3047 1899-12-31 09:35:04 7.3048 1899-12-31 09:35:35 7.3049 1899-12-31 09:36:04 7.3050 1899-12-31 09:36:35 7.3051 1899-12-31 09:37:04 7.3052 1899-12-31 09:37:35 7.3053 1899-12-31 09:38:04 7.3054 1899-12-31 09:38:35 7.3055 1899-12-31 09:39:04 7.3056 1899-12-31 09:39:35 7.3057 1899-12-31 09:40:04 7.3058 1899-12-31 09:40:35 7.3059 1899-12-31 09:41:04 7.3060 1899-12-31 09:41:35 7.30 plot(var1 ~time     ,df)

 mod<-lm(var1 ~time     ,df) summary(mod)Call:lm(formula = var1 ~ time, data = df)Residuals:      Min        1Q    Median        3Q       Max -0.027749 -0.016887 -0.003292  0.012073  0.085639 Coefficients:              Estimate Std. Error t value Pr(>|t|)    (Intercept) -9.375e+04  1.217e+04  -7.706 1.91e-10 ***time        -4.244e-05  5.507e-06  -7.707 1.90e-10 ***---Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘’ 1Residual standard error: 0.0222 on 58 degrees of freedomMultiple R-squared:  0.5059,    Adjusted R-squared:  0.4974 F-statistic: 59.39 on 1 and 58 DF,  p-value: 1.902e-10

So I removed the rows of constant values of var1 and did the linear model again.

 df2<-df[-c(24:60),] plot(var1 ~time     ,df2)

But the slope of the lm gives NA and no R-squared:

 mod<-lm(var1 ~time     ,df2) summary(mod)Call:lm(formula = var1 ~ time, data = df2)Residuals:     Min       1Q   Median       3Q      Max -0.04087 -0.03087 -0.01087  0.01913  0.09913 Coefficients: (1 not defined because of singularities)            Estimate Std. Error t value Pr(>|t|)    (Intercept) 7.340870   0.008219   893.1   <2e-16 ***time              NA         NA      NA       NA    ---Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘’ 1Residual standard error: 0.03942 on 22 degrees of freedom

There must be an slope, since var1 is changing with time. How can I get the slope?The summary warns about a singularity, but I have just one factor, so where is the correlation?

Thanks in advance!