The distance travelled `d` by the car is the area under the velocity graph from time `t=0` to time `t=6` seconds.

Using the trapezium rule with `m=6` equal-width intervals` `

`d approx h/2[y_0+y_m + 2(y_1+y_2+y_3+y_4+y_5)]`

where `h` is the interval width (=1) and `m=6`

Therefore, reading approximate values of `y_0,y_1,y_2,y_3,y_4,y_5,y_6` off the graph at `x=0,1,2,3,4,5,6` respectively

`d approx 1/2[52.5 + 0 +2(37.5+26.25+18.75+11.25+3.75)]`

`= 1/2(52.5+195) = 123.75` ft

**The car travels 123.75ft as it brakes**

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