BMAT 2015 Q 23 motion in one dimension

A car is being driven at 20m/ s when the driver sees a child run into the road. The driver’s usual reaction time is 0.70s, but this is doubled because the driver is tired. Once the driver applies the brakes, the car is brought uniformly to rest in a further 3.3s. What is the total distance travelled by the car between when the driver first sees the child to when the car stops?

A 33 m
B 40 m
C 47 m
D 61 m
E 66 m
F 80 m
G 94 m

Hi everyone, the answer is E)

Here is my work what did i do wrong?
u = 20 m/s
v = 0 m/s
t = (0,7 * 2) + 3,3 = 4,7 seconds

s = 1/2(u+v)t = 47 m

Everything is great but I think there is an error with the formula.

distance = velocity * time

In this case, v = 20m/s,

will be 3.3s.

20m/s * 3.3s = 66m.

It’s great to cancel out some information that is added just to confuse students, there is no need to consider the initial/final velocity or reaction time of the driver.

Thank you it makes sense now, the change in speed only happens after the 1,4s reaction time so over the course of 3,3 seconds