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qaws [65]
3 years ago
9

A vehicle has an initial velocity of v0 when a tree falls on the roadway a distance xf in front of the vehicle. The driver has a

perception-reaction time of t. When the driver begins braking, the vehicle has an acceleration of magnitude a. Write an expression for the velocity of the vehicle when it hits the tree.
Physics
1 answer:
Korvikt [17]3 years ago
5 0

Answer:

v^2=v_o^2-2\times a\times (v_o.t)

Explanation:

Given:

Initial velocity of the vehicle, v_o

distance between the car and the tree, x_f

time taken to respond to the situation, t

acceleration of the car after braking, a

Using equation of motion:

v^2=u^2+2a.s ..............(1)

where:

v= final velocity of the car when it hits the tree

u= initial velocity of the  car when the tree falls

a= acceleration after the brakes are applied

s= distance between the tree and the car after the brakes are applied.

s=v_o\times t

Now for this situation the eq. (1) becomes:

v^2=v_o^2-2\times a\times (v_o.t) (negative sign is for the deceleration after the brake is applied to the car.)

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4 0
3 years ago
A 97.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.63 r
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Answer:

the final angular velocity of the platform with its load is 1.0356 rad/s

Explanation:

Given that;

mass of circular platform m = 97.1 kg

Initial angular velocity of platform ω₀ = 1.63 rad/s

mass of banana m_{b} = 8.97 kg

at distance r = 4/5  { radius of platform }

mass of monkey m_{m} = 22.1 kg

at edge = R

R = 1.73 m

now since there is No external Torque

Angular momentum will be conserved, so;

mR²/2 × ω₀ = [ mR²/2 + m_{b} (\frac{4}{5} R)² + m_{m}R² ]w

m/2 × ω₀ = [ m/2 + m_{b} (\frac{4}{5} )² + m_{m} ]w

we substitute

w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1

w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )

w = 48.55 × [ 1.63 / ( 76.3908 ) ]

w = 48.55 × 0.02133

w = 1.0356 rad/s

Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s

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