A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately pursues it, accelerating at a rate of 10 mi/hr per second. The road is fairly busy, so the officer will not go faster than a top speed of 70 mi/hr. How longwill it take before the officer catches up to the speeding car, and how far will it have travelled inorder to do so

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

 = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

so

=   + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁  = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁  will be

= × t₁

we substitute

= 22.352 × 7

 = 156.46 m

now distance between polive car and speeding car

Δd =   - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( - )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b)  how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

specific latent heat of fusion, lf, of a substance is defined as the amount of heat required to change a unit mass of the substance from solid to liquid state, without any change in the temperature.

explanation:

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