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2 years ago

What average net force is required to stop a 1950 kg car in 10.5 s if it’s initially traveling at 28m/s

Physics

1 answer:

nika2105 [10]2 years ago

7 0

Answer:

<em>An average net force of 5200 N is needed to stop the car</em>

Explanation:

<u>Cinematics and Dynamics</u>

Cinematics describes the variables involved in the movement without dealing with its causes. There are four main concepts in cinematics: Velocity (or its scalar equivalent, the speed), acceleration, time, and displacement (or the scalar equivalent, distance).

The acceleration can be calculated by:

$\displaystyle a=\frac{v_f-v_o}{t}$

The initial speed is vo=28 m/s, it stops (vf=0) in t=10.5 seconds, thus the acceleration is:

$\displaystyle a=\frac{0-28}{10.5}$

$a = -2.67~m/s^2$

The acceleration is negative because the car loses speed.

Knowing the mass of the car m=1950 Kg, we can calculate the net force required to stop the car by using the formula:

F = m.a =1950*2.67

F = 5200 N

An average net force of 5200 N is needed to stop the car

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3 years ago

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1 year ago

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

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Answer:

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Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have: