Recall this kinematic equation:
a = 
This equation gives the acceleration of the object assuming it IS constant (the velocity changes at a uniform rate).
a is the acceleration.
Vi is the initial velocity.
Vf is the final velocity.
Δt is the amount of elapsed time.
Given values:
Vi = 0 m/s (the car starts at rest).
Vf = 25 m/s.
Δt = 10 s
Substitute the terms in the equation with the given values and solve for a:
a = 
<h3>a = 2.5 m/s²</h3>
Kinetic energy is the energy for a catapult.
Answer: a) - 437.8° F, b) - 261°c.
Explanation: a) the kelvin and Fahrenheit temperature scale are related by the formulae below.
5 (°F - 32) = 9 (k - 273)
Where °F = temperature in Fahrenheit and k = temperature in kelvin.
For question A, k = 12.0, by substituting to have the value for °F, we have
5(°F - 32) = 9 ( 12 - 273)
5(°F - 32) = 9(-261)
5(°F - 32) = - 2349
°F - 32 = - 2349/5
°F - 32 = - 469.8
°F = - 469.8 + 32
°F = - 437.8
Question B
The centigrade and kelvin scale are related by the formulae below
°c = k - 273
Where °c = temperature in centigrade and k = temperature in kelvin =12
°c = 12 - 273
°c = - 261
Since it is dropped, initial velocity u = 0
Using s = ut + (1/2)gt^2, putting u = 0, g = 10 m/s^2
s =(1/2)gt^2, t = 3s
s = 0.5 * 10 * 3 *3
s = 45 m.
The bridge is 45m above the water.
