A car is travelling at 2 0m/s when it pulls out to pass a truck that is travelling at only 18 m/s the car accelerates at 2 m/s square for 4 sec and then maintains this new velocity. if the car was originally 8 mtrs behind the truck when it pulled out to pass , how far in front of the truck is the car 10 sec later
Vf = Vo + a*t = 20 + 2*4 = 28 m/s.
Dc = 20*4 + 0.5*2*4^2 = 96 m. in 4 sec.
Dc = 28m/s * 6s = 168 m. in 6 sec.
Dc = 96 + 168 = 264 m. in 10 sec.
Dt = 18m/s * 10s. = 180 m. in 10 sec.
(264-8) - 180 = Distance in front of the truck.
To find the distance in front of the truck that the car is after 10 seconds, we need to calculate the distance traveled during two stages: the passing maneuver and the subsequent motion at constant velocity.
Stage 1: Passing maneuver
1. Calculate the time it takes for the car to catch up to the velocity of the truck:
- The car is initially traveling at 20 m/s, while the truck is traveling at 18 m/s.
- The car's acceleration is 2 m/s².
- To find the time it takes for the car to catch up, we need to calculate the time it takes for the car to reach the truck's initial velocity (18 m/s) from its own initial velocity (20 m/s).
Use the formula: v = u + at, where:
v = final velocity (18 m/s)
u = initial velocity (20 m/s)
a = acceleration (2 m/s²)
t = time
Rearranging the formula to solve for t:
t = (v - u) / a
Substituting the values:
t = (18 - 20) / 2
t = -2 / 2
t = -1 second
Since the time cannot be negative, it means that the car has already caught up to the truck's velocity at the start of the scenario.
2. Calculate the distance covered during the passing maneuver:
- The car is initially 8 meters behind the truck when the passing maneuver begins.
- We need to calculate the distance covered by the car during the acceleration phase of 4 seconds.
Use the formula: s = ut + 0.5at², where:
s = distance
u = initial velocity (20 m/s)
t = time (4 seconds)
a = acceleration (2 m/s²)
Substituting the values:
s = (20 * 4) + 0.5 * 2 * (4)^2
s = 80 + 0.5 * 2 * 16
s = 80 + 16
s = 96 meters
Since the car started 8 meters behind the truck and has moved 96 meters forward in the acceleration phase, the car is now 96 + 8 = 104 meters in front of the truck at the end of stage 1.
Stage 2: Motion at constant velocity
After the acceleration phase, the car maintains a constant velocity of 18 m/s, the same as the truck. The time for this stage is given as 10 seconds.
Since the velocity is constant, the distance traveled during this stage can be calculated using the formula: s = vt, where v = velocity (18 m/s) and t = time (10 seconds).
Substituting the values:
s = 18 * 10
s = 180 meters
Therefore, the car is 180 meters in front of the truck after 10 seconds.
Note: In this scenario, the calculations are based on the assumption that the truck maintains a constant velocity during the entire duration.