Question

Question 4. (7 Marks) I A car and a truck start from rest at the same instant, with the car initially at some 'x' distance behind the truck. The truck was given a constant acceleration of 2.27 ms2 and at the same moment the car was given an acceleration of 5.5 ms 2. The car overtakes the truck after the back of the truck has moved 41.0 m in total. As it is show in the diagram below, the truck is 20 m long and the car is 7 m long. i. How much time does it take for the car to overtake the truck? (2 Marks) ii. How far was the car behind the truck initially? (3 Marks) iii. What is the final velocity of the car? (2 marks)

Answer 1

part 1

Consider that the truck is at the origin at rest. When the car overtakes the truck wh have x = 41 m and v₀ = 0

so

$s = v_0t+\frac{1}{2}a_t t^2$

$41 = 0+\frac{1}{2}(2.27) t^2$

This is the car takes to overtake the truck.

part ii

The car is at 41 m after time 6.01 s so,

$s -s_0= v_0t+\frac{1}{2}a_c t^2$

$41 -s_0= 0+\frac{1}{2}(5.5) (6.01)^2$

So the car was 58.33 m behind the truck initially.

Part iii

Final velocity of car