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Answer :
From the calculations and the data supplied, the braking distance is 65 m
What is the braking force?
The braking force is the force that is applied as the object comes to rest.
We know that the mass of the car is obtained from; 14,700 N/10 m/s^2 = 1470 Kg
Now;
a = F/m = 7100 N/1470 Kg
a = 4.8 m/s^2
Given that;
v^2 = u^2 - 2as
u^2 = 2as
s = u^2/2a
s = (25)^2/2 * 4.8
s = 625/9.6
s = 65 m
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vo = 25 m/sec
vf = 0 m/sec
Fμ = 7100 N (Force due to friction)
Fg = 14700 N
With the force due to gravity, you can find the mass of the car:
F = ma
14700 N = m (9.8 m/sec²)
m = 1500 kg
Now, we can use the equation again to find the deacceleration due to friction:
F = ma
7100 N = (1500 kg) a
a = 4.73333333333 m/sec²
And now, we can use a velocity formula to find the distance traveled:
vf² = vo² + 2a∆d
0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d
0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d
-625 m²/sec² = (-9.466666666667 m/sec²) ∆d
∆d = 66.0211267605634 m
∆d = 66.02 m
vo = 25 m/sec
vf = 0 m/sec
Fμ = 7100 N (Force due to friction)
Fg = 14700 N
With the force due to gravity, you can find the mass of the car:
F = ma
14700 N = m (9.8 m/sec²)
m = 1500 kg
Now, we can use the equation again to find the deacceleration due to friction:
F = ma
7100 N = (1500 kg) a
a = 4.73333333333 m/sec²
And now, we can use a velocity formula to find the distance traveled:
vf² = vo² + 2a∆d
0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d
0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d
-625 m²/sec² = (-9.466666666667 m/sec²) ∆d
∆d = 66.0211267605634 m
∆d = 66.02 m
