I have a physics problem, please help?

Question

Mrs. Jones sees an obstruction in the road while driving and takes .8 seconds to react and press the brake. Her car is travelling at 25 meters/s. a. if when the brake is applied, the car decelerates at a uniform rate of 9.3 m/s squared what is the total distance or displacement.

Second:

Object starts from rest and accelerates at 3 meters/second squared for 4 s. Velocity remains constant for 7 s and it comes to rest with uniform deceleration after another 5 s. Find a. the displacement for each stage of the motion and b. the average velocity over the whole time interval.

Answer 1

there are 3 main equations to use when given a question like this and use the appropriate one acording to the given information:

where

s =distance in m

u=initial velocity in m/s

v=final velocity in m/s

a=acceleration in m/s^2

t=time in s

s=ut+1/2at^2

v^2=u^2+2as

v=u+at

Now with these equations, for the first part we will use the first eqution as we are given time t =0.8s, initial velocity u =25m/s and deceleration a=9.3m/s^2

s=ut+1/2at^2

s=(25x0.80)+(0.5x9.3x0.8^2)

=20+2.976

=22.976

=23m

for the second part, :find the distance for the first time interval. given information is initial velocity =0m/s, as it is at rest, acceleration a=3m/s^2 and time t=4s

s=ut+1/2at^2

s=(0x4)+(0.5x3x4^2)

s=0+24

s=24m

finding the final velocity for this same period

v=u+at

v=0+(3x4)

v=12m/s

the second part states the velocity remains CONSTANT; the initial and final velocity  u an v are the same=12m/s, acceleration =0m/s as acceleration is a result is change in velocity, and the time t=7s. we find this distance:

s=ut+1/2at^2

=(12x7)+(0.5x0x7^2)

=84m

THEN, there is the part for the uniform deceleration for 5 seconds. the initial velocity is 12m/s, time t=5s and final velocity =0m/s, as it is AT REST. we need to find the deceleration before the distance, so

a=(v-u)/t

=(0-12)/5

=-(12/5)

=2.4m/s^2

now we can find the distance

s=ut+1/2at^2

=(12x5)+(0.5x2.4x5^2)

=60+30

=90m

therefore the total distance for the second stage =90+84=174m

we found the displacements to be 24m and 174m, so part a is solved. for part b, we use the basic equation for average speed:

average velocity=total displacement/total time

=(24+174)/(4+5+7)

=198/16

=12.375

=12.4m/s is the average velocity

hope it helps, i really tried :)