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Natasha2012 [34]
3 years ago
8

A car of mass m, traveling at speed v, stops in time t when maximum braking force is applied. Assuming the braking force is inde

pendent of mass, what time would be required to stop a car of mass 2m traveling at speed v?
A.½tB. t√C. 2 tD. 2t
Physics
1 answer:
maw [93]3 years ago
7 0

Answer:

option D

Explanation:

given,

mass of the car 1 = m

speed of car 1 = v

time taken to stop the car = t

mass of car 2 = m' = 2 m

speed of the car 2 = v

time taken by the car to stop = t' = ?

now,

we know,

F = m a....(1)

and force by the second car

F = m' a'

F = 2 m a'

a' = \dfrac{F}{2m}

from equation (1)

a' = \dfrac{a}{2}

using equation of motion

v = u + at

0 = v - a t

t = \dfrac{v}{a}

again using equation of motion for the calculation of the time taken by the second car.

t'= \dfrac{v}{a'}

t' = 2\dfrac{v}{a}

t' = 2 t

hence, the time taken by the second car is twice the time taken by the first car to stop.

The correct answer is option D

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3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

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