A loaded truck is going at a speed of 36 km/hr and a car starts at point 3 km behind the truck with an acceleration of 2 m/s. when will the car overtake the car? What is the distance covered by the car before it overtakes the car?
at²/2 - v1•t =sₒ,how it came ?
To find out when the car overtakes the truck and the distance covered by the car, we need to determine the time taken by the car to catch up with the truck.
First, let's convert the speed of the truck from km/hr to m/s:
Speed of the truck = 36 km/hr = (36 * 1000) m / (60 * 60) s = 10 m/s
Now, let's analyze the motion of the car by using the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance covered by the car
u = initial velocity of the car
a = acceleration of the car
t = time taken
We know that the initial distance between the car and the truck is 3 km = 3000 m and the initial velocity of the car is 0 m/s.
So, the equation becomes:
s = (1/2)at^2
To find the time taken by the car to overtake the truck, we'll equate the distance covered by the car (s) to the distance between the car and the truck (3000 m). We'll solve for t.
3000 = (1/2)(2)t^2
Simplifying the equation:
3000 = t^2
t^2 = 3000
t = √3000
t ≈ 54.77 s
Therefore, the car will overtake the truck at approximately 54.77 seconds.
To calculate the distance covered by the car before overtaking the truck, we'll substitute the value of t into the equation:
s = (1/2)(2)(54.77)^2
s ≈ 2998.3 m
Hence, the distance covered by the car before overtaking the truck is approximately 2998.3 meters.
thanx it really ate my head
acceleration has to be 2 m/s²(!!!)
v1 =36 km/h =36000/3600=10 m/s
at²/2 - v1•t =sₒ
2 •t²/2- 10t=3000,
t² -10t-3000=0
t =10±sqrt(100+3000)= 10±55.68.
t=65.68 s.
s= at²/2=2•65.68²/2=4313.6 m.