Respuesta :
a) Mind you before your reaction time, you had be going at a uniform speed 20m/s so for the reaction time of 0.5 seconds, you had covered a distance of:
20m/s*0.5s = 10 m
For the second part which involved deceleration, using:
v = u - at, Noting that there is deceleration.
u = 20m/s, v = final velocity = 0, a = -10m/s².
Let us solve for the time.
v = u + at
0 = 20 - 10*t
10t = 20
t = 20/10 = 2 seconds.
Let us compute for the distance covered during the 2s
s = ut + 1/2at², a = -10 m/s²
s = 20*2 -0.5*10*2² = 20m
So the total distance covered = Distance covered from reaction time + Distance covered from deceleration
= 10m + 20m = 30m
So you have covered 30m out of the initial 35m.
Distance between you and the dear: 35 - 30 = 5m
So you have 5m between you and the deer. So you did not hit the deer.
20m/s*0.5s = 10 m
For the second part which involved deceleration, using:
v = u - at, Noting that there is deceleration.
u = 20m/s, v = final velocity = 0, a = -10m/s².
Let us solve for the time.
v = u + at
0 = 20 - 10*t
10t = 20
t = 20/10 = 2 seconds.
Let us compute for the distance covered during the 2s
s = ut + 1/2at², a = -10 m/s²
s = 20*2 -0.5*10*2² = 20m
So the total distance covered = Distance covered from reaction time + Distance covered from deceleration
= 10m + 20m = 30m
So you have covered 30m out of the initial 35m.
Distance between you and the dear: 35 - 30 = 5m
So you have 5m between you and the deer. So you did not hit the deer.
So the dear is 35m away from the car that is running at a speed of 20m/s where your reaction time in hitting the brakes is 0.5s and the cars deceleration is 10 m/s2. By calculating i came up with an answer of 5meters, the car only have 5 meters away from the deer to hit it.