Lets denote the distance, velocity and time of the first part for bike P as d₁, v₁, t₁, for the second part as d₂, v₂, t₂ and for the third part as d₃, v₃, and t₃.

We can determine the speed of bike T by using the definition of speed;

Speed of bike T = Distance covered by bike T / time taken

v_T = d_T/ t ………….. (1)

Distance covered by bike P in the first part = d₁ = 80 km = 80 x 10³ m

Distance covered by bike P in the second part = d₂ = 50 km = 50 x 10³ m

Distance covered by bike P in the third part = d₃ = 100 km = 100 x 10³ m

So the total distance covered by bike P = d_P = d₁ + d₂ + d₃ = 80 x 10³ m + 50 x 10³ m + 100 x 10³ m = 230 x 103 m

Since both bikes has to cover the same distance So,

Distance covered by bike T = d_T = 230 x 10³ m

Also, both bikes reach the destination at the same time, so time taken by bike P is equal to time taken by bike T.

Time taken by bike P in the first part = t₁ = d₁/v₁ = 80 km / 30 km/h = 2.67 h = 9612 s

Time taken by bike P in the second part = t₂ = d₂/v₂ = 50 km / 45 km/h = 1.11 h = 3996 s

Time taken by bike P in the third part = t₃ = d₃/v₃ = 100 km / 30 km/h = 3.33 h =11988 s

Time taken by bike P = Time taken by bike T =t = t₁ + t₂ + t₃ = 9612 s + 3996 s + 11988 s = 25596 s

Substituting values of d_T and t in eq(1)

v_T = 230 x 10³ m / 25596 s = 8.99 m/s ≈ 9.0 m/s

A rushing ambulance which is carrying an injured patient is approaching the hospital with a distance from the hospital defined by x(t) = bt2 - ct3, where b = 3.10 m/s2 and c = 0.147 m/s3. Determine the average velocity of the ambulance for t = 0 to t = 7.0 s.

A rushing ambulance which is carrying an injured patient is approaching the hospital with a distance from the hospital defined by x(t) = bt² - ct³, where b = 3.10 m/s² and c = 0.147 m/s³. Determine the average velocity of the ambulance for t = 0 to t = 7.0 s.