You"re driving dvery own the highmeans late one night at 16 m/s once a deer measures onto the road 35 m in front of you. Your reactivity time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 12 m/s2.

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How a lot distance is in between you and the deer when you involved a stop?

What is the maximum speed you could have and still not hit the deer?

Answer

Substitute 16 mathrm~m / mathrms for v, and also 0.50 mathrm~s for t in the equation x=v t and also calculate x.

eginalignedx=(16 mathrm~m / mathrms)(0.50 mathrm~s) \=8.0 mathrm~mendalignedReararray the kinematics equation of activity v_mathrmf^2=v_mathrmi^2+2 a(Delta x) to fix for Delta x.

eginalignedv_mathrmf^2=v_mathrmi^2+2 a(Delta x) \2 a(Delta x)=v_mathrmf^2-v_mathrmi^2 \Delta x=fracv_mathrmf^2-v_mathrmi^22 aendalignedSubstitute 0 for v_mathrmf, 16 mathrm~m / mathrms for v_mathrmi, and -12 mathrm~m / mathrms^2 for a in the over equation Delta x=fracv_mathrmf^2-v_mathrmi^22 a.

eginalignedDelta x =frac0^2-(16 mathrm~m / mathrms)^22left(-12 mathrm~m / mathrms^2 ight) \=frac-256 mathrm~m^2 / mathrms^2-24 mathrm~m / mathrms^2 \=10.67 mathrm~mendalignedExplanatlon

The distance travelled by a pshort article at constant rate is, x=v t

Here, v is the speed, and also t is the time took a trip. The kinematics equation of movement is,

v_mathrmf^2=v_mathrmi^2+2 a(Delta x)

Here, v_mathrmf is the final rate, v_mathrmi is the initial rate, a is the acceleration, and $Delta x$ is the displacement of the body. The car pertains to remainder after decelerating so the last speed is zero and also deceleration applies that a will be negative.

The distance in between the vehicle and also deer when vehicle involves a stop is,

d=L-x-Delta x

Here, d is the distance in between the auto and deer, L is the distance between the automobile and deer at the moment deer measures on the rod, x is the distance vehicle travelled throughout the reaction time, and also Delta x is the distance car travelled after brakes are used.

Substitute 35.0 mathrm~m for L, 8.0 mathrm~m for x, and also 10.67 mathrm~m for Delta x in the equation d=L-x-Delta x and calculate d.

d=35.0 mathrm~m-8.0 mathrm~m-10.67 mathrm~m

=16.3 .3 mathrm~m

The distance in between the car and also deer when vehicle pertains to a speak is 16.33 mathrm~m.

Explanatlon

The distance in between the automobile and also deer as soon as the auto stops is the difference of the initial distance in between the car and also deer, and the distance automobile took a trip during the reaction time and also the distance car took a trip after hitting the brakes.

d=L-x-Delta x

Here, d is the distance between the automobile and also deer, L is the distance in between the vehicle and deer at the moment deer actions on the rod, x is the distance vehicle took a trip in the time of the reaction time, and also Delta x is the distance vehicle travelled after brakes are used.

The distance that car can travel prior to hitting the deer is,

d=L-x

Substitute $35.0 mathrm~m$ for $L$, and 8.0 mathrm~m for x in the equation d=L-x and calculate d.

eginalignedd=35.0 mathrm~m-8.0 mathrm~m \=27.0 mathrm~mendalignedUse the kinematics equation of movement v_mathrmf^2=v_mathrmi^2+2 a(Delta x) and resolve for initial speed.

eginalignedv_mathrmf^2=v_mathrmi^2+2 a(Delta x)\v_i^2=v_mathrmf^2-2 a(Delta x)\v_mathrmi=sqrtv_mathrmf^2-2 a(Delta x)endalignedSubstitute 0 for v_mathrmf,-12 mathrm~m / mathrms^2 for a, and 27.0 mathrm~m for Delta x in the above equation v_mathrmi=sqrtv_mathrmf^2-2 a(Delta x) and calculate v_mathrmi.

eginalignedv_mathrmi=sqrt(0)^2-2left(-12 mathrm~m / mathrms^2 ight)(27.0 mathrm~m)=sqrt648 mathrm~m^2 / mathrms^2 \=25.46 mathrm~m / mathrmsendalignedThe maximum speed that vehicle can have actually and also still not hit the deer is 25.46 mathrm~m / mathrms.

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Explanatlon

The kinematics equation of motion is,

v_mathrmf^2=v_mathrmi^2+2 a(Delta x)

Here, v_mathrmf is the final rate, v_mathrmi is the initial speed, a is the acceleration, and Delta x is the displacement of the body. The distance between the automobile and also deer as soon as the automobile stops is zero to resolve for maximum rate the car can travel. The difference of the initial distance between the vehicle and also deer, and the distance auto took a trip in the time of the reactivity time is the distance vehicle deserve to travel prior to hitting the deer.