Answer:
The truck’s speed, relative to the ground = 38.25 ms⁻¹
Explanation:
As per the question,
Given data:
Velocity of car with respect to ground = 29 ms⁻¹
Velocity of truck with respect to car = 48 ms⁻¹
so
now
Using the vector sum property in the vector triangle, we can say
[tex]V_{t,g}=V_{t,c}+V_{c,g}[/tex]
As we have given the magnitude of velocities
Now,
By using Pythagoras theorem in triangle AOB
OA² + OB² = AB²
OA = √(48² - 29²) = 38.25 ms⁻¹
Therefore, the truck’s speed, relative to the ground = 38.25 ms⁻¹