7. A police officer observes a car approaching at a
uniform speed of 60 km/h. He, on his
motorbike, starts chasing the car as it just crosses
him. After accelerating at a constant
rate of 10 s, he attains his top speed of 75 km/h.
How long does it take to overtake the car
and at what distance?
Answers
Answer:
45 m distance................
Concept:
The second equation of motion is defined as,
s = ut + (1/2) at²
where s is the distance traveled, u is the initial velocity, a is acceleration and t is the time.
Given:
The uniform speed of the car, v₁ = 60 km/h = 16.67 m/s
The speed of the bike after accelerating, v = 75 km/h = 20.84 m/s
Find:
The height from the ground and the time at which both stones will meet.
Solution:
According to the Equation of motion,
v = u+ at
The acceleration of the bike can be calculated by v = u+at,
as at t = 10 s, v = 20.84 m/s and initial velociy is 0.
20.84 = 0+ a (10)
a = 2.084 m/s²
Distance traveled by car,
s = ut + (1/2) at²
As the speed is uniform, the acceleration is equal to zero.
s = 16.67 t
Similarly, the distance traveled by bike,
Initially, the bike is accelerating,
s₁ = ut + (1/2) at₁ ²
Initially, the initial velocity is zero, u=0,
s₁ = 0t + (1/2) a(10) ² = (1/2) ×2.084 (10) ² = 104.2 m
After that the speed is constant,
s₂ = ut₂ + (1/2) at₂²
As the speed is uniform, the acceleration is equal to zero.
s₂ = 20.84 t₂
As the police overtake the car,
s = s₁ + s₂ and t = t₁ + t₂ = 10 + t₂, t₂ = t - 10
The value of s₂ = s - s₁ = s - 104.2 m and s₂ = 20.84 t₂,
So, s - 104.2 = 20.84 t₂
s = 104.2 + 20.84 t₂
16.67 t = 104.2 + 20.84 ( t -10)
4.71 t = 96.64
t = 96.64/4.71 = 20.51 seconds.
At t = 20.51 seconds, distance traveled,
s = 16.67 t = 16.67× 20.51 = 341.90 m
Hence, the police officer will overtake at a distance of 341.90 m after 20.51 s.
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