A car is traveling at a 20.0 m/s for 7.00 s and then suddenly comes to a stop over a 3 s period.

Physics

1 answer:

Delicious77 [7]2 years ago

7 0

Answer: A) Deceleration of the car is -6.6667 m/s² while it came to stop.

B) The total distance the car travels is 200 meter during the 10 s period.

Explanation:

Given Data

Initial velocity of the car ( $v_{i}$ ) = 20.0 m/s

Final velocity of the car ( $v_{f}$ ) = 0 m/s

Time (in motion) =7.00 s

Time (in rest) =3 s

To find - A) car's deceleration while it came to a stop

B) the total distance the car travels in 10 s

A) The formula to find the deceleration is

Deceleration = (( final velocity - initial velocity ) ÷ Time) (m/s²)

Deceleration = (( $v_{f}$ ) - ( $v_{i}$ )) ÷ time (m/s²)

Deceleration = ( 0.0 - 20 ) ÷ 3 (m/s²)

Deceleration = (- 20) ÷ 3 (m/s²)

Deceleration = - 6.6667 m/s²

(NOTE : Deceleration is the opposite of acceleration so the final result must have the negative sign)

The car's deceleration is - 6.6667 m/s² while it came to a stop

B) The formula to find the distance traveled by the car is

Distance traveled by the car is equals to the product of the speed and time

Distance = Speed × Time (meter)

Distance = 20.0 × 10

Distance = 200 meters

The total distance the car travels during the period of 10 s is 200 meters

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I was driving along at 20 m/s, trying to change a CD and not watching where I was going. When I looked up, I found myself 45 m f

cricket20 [7]

Answer:

$a=4.44\frac{m}{s^2}$

Explanation:

First we have to find the time required for train to travel 60 meters and impact the car, this is an uniform linear motion:

$t=\frac{d}{v}\\\\t=\frac{60m}{30\frac{m}{s}}=2s$

The reaction time of the driver before starting to accelerate was 0.50 seconds. So, remaining time for driver is 1.5 seconds.

Now, we have to calculate the distance traveled for the driver in this 0.5 seconds before he start to accelerate. Again, is an uniform linear motion:

$d=vt\\d=20\frac{m}{s}(0.5s)=10m$

The driver cover 10 meters in this 0.5 seconds. So, the remaining distance to be cover in 1.5 seconds by the driver are 35 meters. We calculate the minimum acceleration required by the car in order to cross the tracks before the train arrive, Since this is an uniformly accelerated motion, we use the following equation:

$d=v_0t+\frac{1}{2}at^2\\a=\frac{2(d-v_0t)}{t^2}\\a=\frac{2(35m-20\frac{m}{s}*1.5s}{(1.5s)^2}\\a=4.44\frac{m}{s^2}$