Answer:19.5 m
Explanation:
Given
coefficient of kinetic Friction
Initial speed
Friction is present so it tries to stop to the object and stops it completely after moving certain distance let say s
maximum deceleration provided by friction is
using equation of motion
where
Answer:
(A) Distance taken by automobile to overtake truck is 175.23 m
(B) Final speed of the automobile is 29.6 m/s.
Explanation:
According to the problem, the automobile is at rest, so its initial speed is zero and the automobile is moving with constant acceleration.
The general equation of motion is:
d = ut + 0.5at² .....(1)
Here d is distance, a is acceleration, t is time and u is initial speed.
The distance (d₁) travel by the automobile after green light in time t is given by the relation :
d₁ = u₁t + 0.5at²
We know that its initial velocity is zero i.e. u = 0 m/s.
d₁ = 0.5at² ......(2)
The truck is moving with constant speed, so its acceleration is zero. Therefore, the distance covered by the truck using equation (1) is;
d₂ = u₂t .......(3)
Since, the distance travelled by automobile and truck are same. Hence,
d₁ = d₂
0.5at² = u₂t
t = 2u₂/a
Substitute 14.8 m/s for u₂ and 2.50 m/s² for a in the above equation.
t =
t = 11.84 s
(A) Distance travelled by automobile, d = u₂t =14.8 x 11.84 = 175.23 m
(B) Final speed of the automobile is given by the relation:
v = u + at
v = 0 + 2.50 x 11.84
v = 29.6 m/s
(a)The required speed for take off will be 24.49 m/sec.
(b)The minimum length must the runway have 193 m.
<h3>What is speed?Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while v for the final speed. its SI unit is m/sec.
(a)The required speed for take off will be 24.49 m/sec.
From the Newtons third equation of motion;
Hence, the required speed for take off will be 24.49 m/sec.
(b)The minimum length must the runway have 193 m.
From the Newtons third equation of motion;
Hence, the minimum length must the runway have 193 m.
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