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timama [110]
3 years ago
8

A car starts to move from rest and covers a distance of 360m in one minute. Calculate the acceleration of the car.

Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
4 0
<h2>The acceleration of car is 0.2 ms⁻²</h2>

Explanation:

When the car moves , the distance covered is calculated by the relation

S = u t + \frac{1}{2} a t²

In this question u = 0 , because car was at rest initially

Thus S =  \frac{1}{2} a t²

here S is displacement and a is the acceleration of car

Therefore  360 =  \frac{1}{2} a ( 60 )²

Because time taken is one minute or 60 seconds

Therefore a = \frac{360x2}{3600}

or a = 0.2 m s⁻²

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A spherical bowling ball with mass m = 3.6 kg and radius R = 0.101 m is thrown down the lane with an initial speed of v = 8.7 m/
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Answer:

1)  The magnitude of the angular acceleration = 67.92 rad/s^{2}

2) Magnitude of the linear acceleration = 2.744 m/s^{2}

3) How long does it take the bowling ball to begin rolling without slipping = 0.906 s

4) How long does it take the bowling ball to begin rolling without slipping = 6.75 m

5) the final velocity is 6.21 m/s

Explanation:

the given information :

Bowling mass m = 3.6 kg

Radius = 0.101 m

Initial speed v_{0} = 8.7 m/s

Coefficient of kinetic friction μ = 0.28

1) he magnitude of the angular acceleration of the bowling ball is

F = m a

F_{g}  = μ N  ,   N = m g

F_{g}  = μ m g

1) The magnitude of the angular acceleration of the bowling ball as it slides down the lane:

momen inersia of Bowling ball I = (2/5) m R^{2}

torque τ = I α

τ = F R

I α = F R

(2/5) m R^{2}  α = μ m g R

α = (5 μ g / 2R) μ g R

  = (5 x 0.28 x 9.8/ 2 x 0.101)

  = 67.92 rad/s^{2}

2) Magnitude of the linear acceleration of the bowling ball as it slides down the lane

F = - F_{g} , F_{k} is the force of kinetic friction

m a = - μ m g, remove m

the magnitude of linear accelaration is

a = μ g

  = (0.28) (9.8)

  = 2.744 m/s^{2}

3) The bowling ball takes time to begin rolling without slipping:

The linear speed, v_{t} = v_{0} - a t

                            v_{t}  =  v_{0} - μ g t

the angular speed, ω = ω0 + α t

                                ω = ω0 + (5  μ g/2R ) t

v_{t} = ω R

v_{0} - μ g t = ω0 R + (5  μ g/2R ) t R

7 μ g t/2 = v_{0} + ω0 R

hence,

t = (2 v_{0} + ω0 R)/  7 μ g

ω0 = 0 (no initial spin), therefore

t = 2 v_{0} / 7 μ g

 = 2 x 8.7 / 7 (0.28) (9.8)

 = 0.906 s

4) How long it takes for the bowling ball to begin rolling without slipping, S

S = v_{0}  t - (1/2) a t^{2}

  = (8.7) (0.906) - (1/2) (2.744) 0.906^{2}

  = 6.75 m

5) The final velocity

v_{t} = v_{0} - a t

v_{t} = 8.7 - (2.744) (0.906)

v_{t} = 6.21 m/s

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