A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a constant acceleration of 2.10 m/s2 and the car an acceleration of 3.40 m/s2. The automobile overtakes the truck after the truck has moved 60.0 m.Requried:a. How much time does it take the car to overtake the truck? b. How far was the car behind the truck initially? c. What is the speed of each when they are abreast? d. On a single graph, sketch the position of each vehicle as a function of time. Take x = 0 at the initial location of the truck.

This is an example of relative motion between two moving bodies and equation of motion are usually modified to solve problems involving relative motion.

In this question, we'll be making use of

x - x₁ = v₁t + ½at²

The above equation is a modification of

x = vt + ½at²

Where x = distance

v = velocity of the body

a = acceleration of the body

t = time

x - x₁ = v₁t + ½at²

Assuming the starting point of the bodies x₁ = 0 which is at rest, the car sped past the truck at 60m.

x - x₁ = v₁t + ½at²

x₁ = 0

v₁ = 0

x = 60

a = 2.10

t = ?

60 - 0 = 0 × t = ½ × 2.10t²

Solve for t

60 = 1.05t²

t² = 60 / 1.05

t² = 57.14

t = √(57.14)

t = 7.56s

The time it took the car to over take the truck is 7.56s

b).

From our previous equation,

x - x₁ = v₁t + ½at²

Same variable except

a = 3.40m/s² and t = 7.56s

60 - x₁ = 0 + ½ × 3.40 × 7.56²

60 - x₁ = 97.16

x₁ = 60 - 97.16

x₁ = -37.16m

The car was behind the truck by 37.16m or the truck was ahead of the car by 37.16m.

c).

The initial speed of the car ?

v = u + at

u = 0m/s

a = 3.40m/s²

v = 0 + 3.40 × 7.56

v = 25.704m/s

The initial speed of the truck = ?

v = u + at

u = 0m/s

a = 2.10m/s

v = 0 + 2.10 × 7.56

v = 15.876m/s

d).

Due to some circumstance, kindly check attached document for a sketch of the graph