A car travelling at 95 km/h strikes a tree

Video Transcript

The initial speed of a car is 95 km per hour. Now, I know we're gonna have to convert this two m per second, so there's 1000 m in a kilometer and there's 30 600 seconds in an hour. Okay, the front of the car only travels a distance of 8/10 of a meter. What's the acceleration during the collision? Well, V Squared final, which is zero. It's going to equal V squared -2 A. D. So, uh, D no, no, no A is going to be V squared over two D. So putting that in a calculator, V equals 95 times 1000 divided by 3600 D. Equals 36000.8 V squared over D Is going to be 870 meters per second squared. Now, if I divide that by G, which is 9.81, then I get 89 Gs. Yeah, 89. Thank you for watching.

Video Transcript

so we have of the initial of ninety five kilometers per hour and we can convert and say there's gonna be times one thousand meters per kilometer and then this will be times one hour for every thirty six hundred seconds and we're getting that. This in meters per second is going to be twenty six point three eight nine meters per second at this time. At this point, we know the Delta X. So the cars is the car is basically being crushed and it comes to a full stuff in point eight zero meters and we need to find the magnitude of the acceleration. So we're going to use schematics. We can say that velocity final squared equals velocity. Initial squared, plus two times acceleration times Delta X. We know that the VF final squared is going to be zero because it's coming to a stop and then we know that a will simply be equal to negative the initial squared over two times out the axe and it's going to be able to negative twenty six point three eight nine squared, divided by two times point eight zero and we're getting that acceleration is going to be equal to negative four hundred and thirty five point two meters per second squared. All we have to do is do a devout of I g in order to find how many g's and we're finding that the ah absolute value of the acceleration is going to be equal to approximately forty four. Jeez. So these will be your two answers acceleration and meters per second square and acceleration in G's. That is our final. That is the end of the solution. Thank you for watching.

Question:

A car traveling 95 km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80 m. What was the magnitude of the average acceleration of the driver during the collision? Express the answer in terms of "g's," where 1.00g = 9.80 {eq}m/s^2 {/eq}.

Accelerated Motion

If the speed of the object changes with time the motion of the object is an accelerated one. If the speed of the object increases with time the acceleration is taken as positive and if the speed decreases with time the acceleration is taken as negative. If we know the initial speed u and final speed v and distance traveled s while accelerating to the final speed, we get the acceleration a of the object from the equation {eq}v^2 = u^2 \pm 2 a s {/eq}. The sign of the acceleration depends on the speed is increasing or decreasing.

Answer and Explanation: 1

Given data

• Speed of the car just before collision {eq}u = 95 \ km/h = 26.39 \ m/s {/eq}

• The driver of the car comes to rest after traveling a distance {eq}d = 0.80 \ m {/eq} on collision.

• Acceleration due to gravity {eq}g = 9.8 \ m/s^2 {/eq}

The driver comes to rest on collision , so the final speed v = 0

Let a be the acceleration of the driver while stopping.

Then we can express the final speed of the car and driver as {eq}0 = u^2 - 2 a d {/eq}

The negative sign has taken for acceleration because the speed is decreasing.

So, the acceleration of the driver {eq}a = \dfrac { u^2 } { 2 d } \\ a = \dfrac{ 26.39^2 } { 2 \times 0.80 } \\ a = 435.27 \ m/s^2 {/eq}

Taking the ratio of acceleration with acceleration due to gravity {eq}\dfrac { a } { g } = \dfrac { 435.27 } { 9.80 } = 44.42 {/eq}

So, the acceleration of the driver while stopping {eq}a = 44.42 \ g {/eq}

Learn more about this topic:

Solving Kinematics Problems

from

Chapter 2 / Lesson 21

Kinematics problems analyze classical mechanics of motion using four standard mathematical equations. Learn the use of these equations to solve kinematics problems using math.

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A car traveling at 120 km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 1.0 m. What was the average acceleration of the driver during the collision? Express the answer in terms of "g's", where 1.00 g = 9.80 m/s2.

Solution: