A 745 i BMW car can brake to a stop in a distance of 121 ft . from a speed of 60.0 mi / h . To b

A 745 i BMW car can brake to a stop in a distance of 121 ft...

A 745 i BMW car can brake to a stop in a distance of 121 $\mathrm{ft}$ . from a speed of 60.0 $\mathrm{mi} / \mathrm{h}$ . To brake to a stop from a speed of 80.0 $\mathrm{mi} / \mathrm{h}$ requires a stopping distance of 211 $\mathrm{ft}$ . What is the average braking acceleration for (a) 60 $\mathrm{mi} / \mathrm{h}$ to rest, (b) 80 $\mathrm{mi} / \mathrm{h}$ to rest, (c) 80 $\mathrm{mi} / \mathrm{h}$ to 60 $\mathrm{mi} / \mathrm{h}$ ? Ex- press the answers in $\mathrm{mi} / \mathrm{h} / \mathrm{s}$ and in $\mathrm{m} / \mathrm{s}^{2}$ .

Key Concepts

Speed Differential Deceleration

This concept involves scenarios where a vehicle decelerates to a speed lower than its initial speed, rather than to a complete stop. It requires adjusting the kinematic equations to account for the reduced change in velocity, illustrating how deceleration is affected by the difference in initial and final speeds.

Average Braking Acceleration

Average braking acceleration is the estimate of a vehicle's deceleration rate calculated over a specific distance and change in speed. It represents the uniform rate at which the vehicle slows down and is obtained by rearranging the kinematic equations to relate initial or final speeds with stopping distances.

Stopping Distance

Stopping distance refers to the total distance a vehicle travels from the moment the driver initiates braking until the vehicle comes to a complete stop. It is directly related to the initial speed of the vehicle and the magnitude of the deceleration, making it an important parameter in safety and performance evaluations.

Unit Conversion

Converting between units is crucial in physics to ensure that calculations are performed consistently. Whether converting speeds from miles per hour to meters per second or distances from feet to meters, correct unit conversions allow standard kinematic formulas to be applied accurately in different measurement systems.

Constant Acceleration Kinematics

This concept uses equations that relate velocity, displacement, acceleration, and time under the assumption of constant acceleration. These kinematic equations are essential in analyzing the motion of objects, such as vehicles braking to a stop or reducing speed, by providing a mathematical framework to relate initial and final velocities with traveled distance and acceleration.

Transcript

00:01 Hi everyone here it is given bmw car can break to stop in a distance of 121 fit from a speed of 60 mile per hour so here given braking distance is given 121 so initial distance is 0 final is 121 for initial speed to be 60 mile per hour and final to be 0 and to break to stop from speed 80 mile per hour to zero distance required is 211 fit so initial distance is zero final is 211 initial speed is 80 mile per hour and final is zero we have to calculate average acceleration for breaking 60 mile per hour speed for breaking 80 mile per hour for breaking 80 mile per hour to 60 mile per hour we have to to express the answer in mile per hour per second as well as in meter per second square.

01:15 I is indicating here initial position f is final position so we can correlate v square final to be v i square initial plus 2a x final minus x initial this relation we will use for first case final velocity is zero initial is 60 mile per hour 2 into a final position is 121 fit minus 0 so from here you will get acceleration to be negative 3 ,600 mile square per hour square upon 1 2 2 into 1 2 into 121 fit so it has to be converted into mile one fit having 5 -280 mile and one hour to be converted into second so it will be 1 upon 360 0 seconds so on solving it this acceleration will be minus 21 .8 mile per hour per second and it can be converted into meter per second square and its value will be minus 9 .75 meter per second square so this is the answer in meter per second square and this is the answer in mile per hour per second this is the answer of a part now b part 0 final velocity is 0 initial velocity is 80 mile per hour plus 2a final distance is 2 to 211 fit minus 0 so acceleration you will get 6400 mile mile square upon h square upon 2 into 211 this will negative now we have to convert it mile square per hour into 3 ,600 second upon 2 into 2 1 1 1 and divided by 1 fit to be converted into mile so we have to divide by just a moment please 5 -280 mile simplifying it you will get the acceleration to be minus 22 .2 mile per hour per second or in meter per second it would be minus 9 .94 meter per second square now in third case initial velocity is 80 mile per hour, final velocity is 60 mile per hour.

06:11 So you write 60 mile per hour square to be initial is 80 mile per hour square plus 2a as s is the distance which will be final minus initial the final distance is 211 fit and initial is 121 fit so from here you will get that's retardation to be minus 280 mile square hour square upon 2 into 90 fit it is to be converted into mile per hour per second to mile square divided by hour into 360 second upon 2 into 90 into 1 upon 5 to 80...

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