Equation of Motion: Calculating Braking Distance on a Straight Road

[nasadall]

Thanks.

This next one is a bit confusing, not so much the first part, its the second, i think i got it wright, i just don't know how to put it on peper to present it to the tutor.

b) A car is being driven along a straight and level road at a steady speed of
25 ms-1 when the driver suddenly notices that there is a fallen tree blocking the
road 65 metres ahead. The driver immediately applies the brakes giving the car
a constant retardation of 5 ms-2.

i) How far in front of the tree does the car come to a halt?

u=25 m/s v=0 m/s a=-5 m/s2 s=?

so using v2=u2+2as

my result was

02=252+2(-5)s
0=625+(-10)s
10s=-0-625
s=-625/-10
s=62.5

then i take away the 62.5 meters treveled while deccelerating, from the 65 meters from where the tree was spoted.

65-62.5 = 2.5 meters that the car came to a halt in front of the tree.

 

[nasadall]

ii) If the driver had not reacted immediately and the brakes were applied one second later, with what speed would the car have hit the tree?


u=25 m/s a=-5 m/s2 s=65 m - (1 sec reaction time = 25 m/s) = 40 m? v=?

same formula

v2=252+(-5)40
v2=625+(-10)40
v2=625+(-400)
v2=225
v=√225
v=15

The car hit the tree with a velocity of 15 m/s