Answer:
The truck’s speed, relative to the ground = 38.25 ms⁻¹
Step-by-step 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
![V_(t,g)=V_(t,c)+V_(c,g)](https://img.qammunity.org/2020/formulas/physics/high-school/ovdp8fgfdmrts6ef50vl2qzqjkwwfqt6wp.png)
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