A car moving with a speed of 90kmh\(^{-1}\) was brought uniformly to rest by the application of the brakes in 10s. How far did the car travel after the brakes were applied?
125m
150m
250m
15km
Explanation
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u=90×1000/36000
u=25ms^-1
a(deceleration)=u/t
a=25/10
a=-2.5
S=ut+½at²
S=25×10+½×-2.5×10²
S=250-125
S=125m
OR
v²=u²+2as
0=25²+2(-2.5)s
625=5s
S=625/5
S=125m
If the velocity time graph is drawn, the distance will be the area of a triangle and not a square. This is because it came to REST. The answer, therefore, should be = 0.5 x 10 x 25 = 125m.
Speed of car = 90km/hr
90*1000
= ------------ => 25m/s
3600
But,
Acceleration = 25/10 = 2.5m/s²
Using
S = ut + ½gt²...
U = 0m/s because the was to rest. Therefore, V is max.
S = 0*10 + ½*2.5*10²
S = 0 + ½*250
S = 125m//
V=90×1000/60×60=25m/s
U=0
t=10s
Recall that
S=v+u(t)/2
S=25+0(10) /2
S= 250/2
S=125m
Here is a better way of answering the question
u= 90km/hr = (90 x 1000)÷ 3600 = 25m/s
v=0 , t= 10, S=?
V = u + at
0 = 25 + 10a
-25 = 10a
a = -25/10 = -2.5
S = ut + 1/2 at²
S = 25 x 10 + 1/2 x (-2.5) x 10²
S = 250 - 125
S = 125m
Using V²=U²+2as
V²=0
U²=25²
a=-2.5
S=?
0=25²+2(-2.5)S
625=5S
S=
625/5=5/5
S=125m.
Speed of car = 90km/hr
90*1000
= ------------ => 25m/s
3600
But,
Acceleration = 25/10 = 2.5m/s²
Using
S = ut + ½gt²...
U = 0m/s because the was to rest. Therefore, V is max.
S = 0*10 + ½*2.5*10²
S = 0 + ½*250
S = 125m//
The question doesn't match the answer
The final velocity is 0 and the initial is 25m/s
