Answer:
It takes the car 9.75 seconds to overtake the truck.
The car starts 92.9 meters at the start.
Step-by-step explanation:
The equation relating distance with initial velocity, acceleration, and time is:
D = vt + 1/2a*t^2, where D is distance, vi is the initial velocity, a is acceleration, and t is time. Acceleration, time, and distance are commonly expressed as m/s^2, sec, and meters.
The truck and car meet at some distance from the start, which we'll call D. D for both the truck and car are equal at this point. Let D' be the distance for the truck at time t and acceleration v1. Let D" be the distance for the car.
Let's find expression for both vehicles:
Truck: D' = vt + 1/2a*t^2
Car: D" = vt + 1/2a*t^2
We are told that the car initially starts behind the truck, some distance we'll call d. So at time 0, the car is at 0 and the truck is at d. (The truck is ahead of the car). We need to add this distance to the equation for the truck.
Truck: D' = vt + 1/2a*t^2 + d
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Let's add information to these two equations from the problem description.
Both start from rest (vt = 0) and we find that the truck has a constant acceleration of 2.1 m/s^2 , while the car has an acceleration of 3.4 m/s2 .
Truck
D' = vt + 1/2a*t^2 + d
D' = 0 + 1/2(2.1 m/s*2*t^2 + d
Car
D" = vt + 1/2a*t^2
D" = 0 + 1/2(3.4 m/s^2)*t^2
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Lets calculate the time the truck travelled to reach 60 meter (D'):
D' = + 1/2a*t^2
60 meters = + 1/2*(2.1 m/s^2)*t^2
(120 meters)/(2.1 m/s^2) = t^2
t^2 = 95.2 seconds
t = 9.75 seconds
This means the car also travelled 9.75 seconds. Let's use the car's equation to find distance:
D" = 0 + 1/2(3.4 m/s^2)*t^2
D" = 0 + 1/2(3.4 m/s^2)*(9.75)^2
D" = (1.7m/s^2)*(95.23s^2)
D" = 161.9 meters.
The truck travelled 60 meters and the car travelled 161.9 meters when they meet. The car therefore started behind the truck by (161.9 - 69) meters.
The car was behind the truck by 92.9 meters at the start.