Just a car starts to accelerate from rest with acceleration 1.4 m/s. A bus moving with constant velocity of 12 m/s passes it in parallel lane.
(a) How long before the car overtakes the bus?
(b) In what velocity will the car than be going?
(c) How far have the car gone at this instant?
bus distance = 12 t
car distance = (1/2) a t^2 = 0.7 t^2
so when
12 t = 0.7 t^2
t = 12/0.7 = 17.1 seconds
v car = a t = 1.4 * 17.1 = 23.9 m/s
distances the same for car and bus = 12 t
after t seconds,
distance of car = 0.7t^2 m
distance of bus = 12t m
a) when is .7t^2 = 12t
.7t^2 - 12t = 0
t(.7t - 12) = 0
t = 0 or t = 12/.7 = 120/7 seconds , obviously rejecting t = 0
b) velocity of car = 1.4t
when t = 120/7 , v = 1.4(120/.7) = 240 m/s (absurd answer! that would be 864 km/h)
c) d = 12(120/7) = 1440/7 m or appr 205 m
What a weird question. That acceleration way out of line, other than
it should say m/s^2. It might have been the case of an publisher just
changing units such as ft/s into metric units to be more current, or else
just making up numbers without thinking.
Anonymous, glad you caught that .
Now the question makes sense, my bad
b) velocity of car = 1.4t
when t = 120/7 , v = 1.4(120/.7) = 240 m/s (absurd answer! that would be 864 km/h)
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tilt, I think you mean 1.4 (120 / 7.0 )
To find the answers to these questions, we can use the equations of motion. Let's break down each question and find the solutions step by step:
(a) How long before the car overtakes the bus?
To find the time it takes for the car to overtake the bus, we need to consider that both vehicles start at the same time, but the bus is already moving with a constant velocity. The car needs to catch up to the bus, so we need to determine when the car's position is equal to the bus's position.
Let's assume the car overtakes the bus after time 't'. The distance covered by the car at time 't' can be calculated using the equation:
distance_car = (1/2) * acceleration * time^2
The distance covered by the bus at time 't' can be calculated using the equation:
distance_bus = velocity_bus * time
Since the car overtakes the bus when they cover the same distance, we can equate these two distances:
(1/2) * acceleration * t^2 = velocity_bus * t
Simplifying the equation, we get:
0.7t^2 = 12t
To find the value of 't', we can rearrange the equation to the following quadratic equation:
0.7t^2 - 12t = 0
Solving this quadratic equation will give us the value of 't', which is the time it takes for the car to overtake the bus.
(b) In what velocity will the car then be going?
To find the velocity of the car when it overtakes the bus, we can use the equation of motion:
final_velocity = initial_velocity + (acceleration * time)
Since the car starts from rest, its initial velocity will be zero. We already know the value of 't' from solving part (a), so we can substitute the values into the equation to find the final velocity of the car.
(c) How far has the car gone at this instant?
To find the distance covered by the car when it overtakes the bus, we can use the equation of motion we used in part (a):
distance_car = (1/2) * acceleration * t^2
Substituting the value of 't' we found in part (a) into this equation will give us the distance covered by the car at that instant.
By following these steps, we can find the solutions for parts (a), (b), and (c). Remember to solve the quadratic equation to find the time 't' in part (a), use the equation of motion to find the final velocity of the car in part (b), and use the equation of motion again to find the distance covered by the car in part (c).