Answer:A)33m
B)32.86m/s
Step-by-step explanation:
Using the equation of motion
V^2=U^2-2aS
Where V=final velocity of the car=0
U=initial velocity=17m/s
a=deceleration=-12m/s^2
S= distance covered by car before stopping
Substituting into the equation
0=17^2-2×12×S
0=289-24S
24S=289
S=289/24
S=12m.
The distance between the car and the deer=45-12=33m
B) Using same equation
Distance,S is now 45m,which is the distance between the car and the deer.while deceleration remains same
V^2=U^2-2aS
0=U^2-2×12×45
U^2=1080
U=√1080
U=32.86m/s
8 + -3x-15/2x^2+×+2 should be the correct answer.
1/5 of a week = 1.4 days
7 / 1.4= 5
5 x 3/4= 3 3/4 novels
AB = 15
CD = 15
Distance from AB to CD = 10
the formula of area K= b*h
K= 15*10 = 150