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balu736 [363]
3 years ago
13

A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and braking system,

and comes to rest 5.5 s later. Find the distance that the car travels during braking
Physics
1 answer:
NARA [144]3 years ago
8 0

As per the question the initial speed of the car [ u] is 42 m/s.

The car applied its brake and comes to rest after 5.5 second.

The final velocity [v] of the car will be zero.

From the equation of kinematics we know that

                                         v=u+at [ here a stands for acceleration]

                                         0=42 +5.5a

                                         a =\frac{-42}{5.5} m/s^2

                                          a= -7.64 m/s^2

Here a is taken negative as it the car is decelerating uniformly.

We are asked to calculate the stopping distance .

From equation of kinematics we know that

                                              S=ut+\frac{1}{2} at^2  [here S is the distance]

                                                      = 42*5.5 +\frac{1}{2} [-7.64] [5.5]^2 m

                                                       =115.445 m      [ans]  

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A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t
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Answers:

a) 65.075 kgm/s

b) 10.526 s

c) 61.82 N

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

I=\Delta p=p_{2}-p_{1} (1)

Where:

I is the impulse

\Delta p is the change in momentum

p_{2}=mV_{2} is the final momentum of the ball with mass m=0.685 kg and final velocity (to the right) V_{2}=57 m/s

p_{1}=mV_{1} is the initial momentum of the ball with initial velocity (to the left) V_{1}=-38 m/s

So:

I=\Delta p=mV_{2}-mV_{1} (2)

I=\Delta p=m(V_{2}-V_{1}) (3)

I=\Delta p=0.685 kg (57 m/s-(-38 m/s)) (4)

I=\Delta p=65.075 kg m/s (5)

<h3>b) Time </h3>

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately 1.0 cm=0.01 m:

V_{2}=V_{1}+at (6)

V_{2}^{2}=V_{1}^{2}+2ad (7)

Where:

a is the acceleration

d=0.01 m is the length the ball was compressed

t is the time

Finding a from (7):

a=\frac{V_{2}^{2}-V_{1}^{2}}{2d} (8)

a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)} (9)

a=90.25 m/s^{2} (10)

Substituting (10) in (6):

57 m/s=-38 m/s+(90.25 m/s^{2})t (11)

Finding t:

t=1.052 s (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force F is proportional to the variation of momentum  \Delta p in time  \Delta t:

F=\frac{\Delta p}{\Delta t} (13)

F=\frac{65.075 kgm/s}{1.052 s} (14)

Finally:

F=61.82 N

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