A 1500 kg car skids to a halt on a wet road where μk = 0.47. You may want to review (Pages 141 - 145) . Part A How fast was the
car traveling if it leaves 68-m-long skid marks?
1 answer:
The car travels at a speed of 25m/s.
<u>Explanation:</u>
Given-
Mass, m = 1500kg
Coefficient of friction, μk = 0.47
Distance, x = 68m
Speed, s = ?
We know,

and
F = μ X m X g
Therefore,
μ * m * g = m * a
μ * g = a
Let, g = 9.8m/s²
So,


We know,

where, v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
If the car comes to rest, the final velocity, v becomes 0.
So,

The car travels at a speed of 25m/s.
You might be interested in
Answer:
acceleration
Explanation:
acceleration =velocity final-velocity initial /time
Work done by the car is 3900 J
Explanation:
- Power and work are related by the equation, Power = Work Done/Time
- Power is the rate at which work is done.
- Here, the car uses power of 260 W and time taken is 15 s.
Work Done = Power × Time
= 260 × 15 = 3900 J
Answer:
uujjjjjctc7tox7txr9ll8rz8lr5xl8r6l8dl85x8rl5x8rl5x8rl5xrx8l58rk5xr8l5xr6l8xr68lc
No it won't. It'll vary inversely as the square of the separation.