Physics problems?

Question

24) in coming to a stop, a car leaves a skid marks of 80 m long on the high way. assuming deceleration of 7 m/s2, estimatae the speed of the car just before preaking?

25) a car traveling at 45 km/h slows down to a constant .5 m/s2 just by letting up the gas. calculate a) the distance of the car coasts before it stops , b) the time it takes to stop and c) the distance it travels during the first and 5th seconds?

I got an answer for the first one, but not sure if its righht, and i dont even know where to start for the second one.

Please also show your work, and how you got your answer, thanks

Answer 1

24) Working formula is

Vf^2 - Vo^2 = 2as

where

Vf = final velocity of the car = 0 (when the car stops)

Vo = initial velocity of the car

a = acceleration = 7 m/sec^2

s = distance travelled = 80 m

Substituting values,

0 - Vo^2 = 2(-7)(80)

NOTE the negative sign attached to "a" as this simply implies that the car was slowing down when the brakes were applied.

Therefore,

Vo^2 = 1120

Vo = 33.47 m/sec.

**************************************...

25a)

Vf^2 - Vo^2 = 2as

where all the terms are the same terms defined in problem 24.

Substituting values,

0 - (45 * 1000/3600)^2 = 2(-.5)s

-156.25 = -1(s)

Solving for "s",

s = 156.25 meters = 0.15625 km.

25b)

The working formula is

Vf - Vo = aT

where

T = time for car to stop

and all the rest of the terms have been previously defined.

Substituting values,

0 - 45 * 1000/3600 = (-0.5)T

Solving for "T",

T = 12.5/0.5

T = 25 seconds

25c)

Working formula is

S = VoT + (1/2)aT^2

At T = 1 sec.

S(1) = (45*1000/3600)(1) + (1/2)(-0.5)(1)

S(1) = 12.5 - 0.25 = 12.25 meters

At T = 5 sec.

S(5) = (45 * 1000/3600)(5) - (1/2)(0.5)(5^2)

S(5) = 62.5 - 6.25

S(5) = 56.25 meters

Answer 2

24)

here s(displacement) is 80m

a(acceleration [+/-]) is -7m/s^2

final velocity 0 m/s

use the equations of motion

u=?, v=0 m/s, a= -7m/s^2, s=80 m

therefore v^2 - u^2 = 2as

so,

       0 - u^2 = 2*(-7)*80

             u^2 = 1120

                  u = 33.466 m/s

25) here u=45 km/h = 12.5 m/s

v=0, a=0.5m/s^2

by equations of motion

v=u+at

therefore t=(v-u)/a

t= -12.5/-0.5

t= 25 sec  (B)

s=ut+0.5at^2

s=12.5*25 - 0.5*0.5*625

s=312.5-156.25 = 156.25 m  (A)

try (C) by this hint

the distance it travels during the first second

will be

s2 - s1

i.e. distance traveled in 2sec - distance traveled in 1sec

the distance it travels during the fifth second

will be

s5 - s4

*********there might be some calculation mistake