Traveling in the city its faster to take a bicycle than a bus, one reason for this is that a bike can accelerate faster than a bus. Let’s say a Bike and bus both stopped at a red light and both takes off an exact same time when the light turns green. The bike accelerates at 1.2 m/s2 with a vmax of 12 m/s . The bus on the other hand has an acceleration of 0.8 m/s2 with a vmax of 20 m.s. The bike has a higher initial acceleration so it will be in front of the bus for some time. How much time elapses until the bus catches up with the biker.

Question

Traveling in the city its faster to take a bicycle than a bus, one reason for this is that a bike can accelerate faster than a bus. Let’s say a Bike and bus both stopped at a red light and both takes off an exact same time when the light turns green. The bike accelerates at 1.2 m/s2 with a vmax of 12 m/s . The bus on the other hand has an acceleration of 0.8 m/s2 with a vmax of 20 m.s. The bike has a higher initial acceleration so it will be in front of the bus for some time. How much time elapses until the bus catches up with the biker.

Abus: 0.8 m/s2
Vmax: 20 m/s
Abike: 1.2 m/s2
Vmax: 12 m/s

Answer 1

a₁=0.8 m/s²

v₁=20 m/s
a₂=1.2 m/s²
v₂=12 m/s
bike: the time to rich v₂ is
t₂ =v₂/a₂ = 12/1.2 = 10 s.
the distance to the meeting point
-for bike
s= a₂t₂²/2 +v₂(t-t₂)
-for bus
s=a₁t²/2
a₂t₂²/2 +v₂(t-t₂)=a₁t²/2
a₁t² -2 v₂t+2 v₂t₂-a₂t₂²=0
0.8 t²-24t+240-120=0
t²-30t+150=0
t=23.7 s

Answer 2

Thank you so much, you hit it right on the nail

Answer 3

To find the time it takes for the bus to catch up with the biker, we can use the equations of motion.

Let's assume that both the bus and the bike start at the same position at time t = 0.

The equations of motion for constant acceleration are:

1) v = u + at, where:
v = final velocity,
u = initial velocity,
a = acceleration, and
t = time.

2) s = ut + (1/2)at^2, where:
s = distance traveled.

Let's first calculate the time it takes for the bike to reach its maximum velocity:

We know that the bike's acceleration is 1.2 m/s^2 and its maximum velocity is 12 m/s.

Using equation 1, we can substitute the given values:
12 m/s = 0 m/s + 1.2 m/s^2 * t_bike

Simplifying the equation:
t_bike = 12 m/s / 1.2 m/s^2
t_bike = 10 seconds

So, it takes the bike 10 seconds to reach its maximum velocity.

Next, let's calculate the distance the bike travels during this time:
Using equation 2:
s_bike = (1/2) * 1.2 m/s^2 * (10 s)^2
s_bike = 60 meters

Now, we can find the time it takes for the bus to catch up with the bike.

To find this time, we need to compare the distances traveled by both the bus and the bike.

Since the bus starts from the same position as the bike and the bike has already traveled 60 meters when the bus starts, we can say that the distance traveled by the bus is the same as the distance traveled by the bike (s_bike).

Using equation 2 with the values for the bus:
s_bus = (1/2) * 0.8 m/s^2 * t_bus^2

Since the bus's maximum velocity is 20 m/s, we can also represent the distance traveled by the bus as:
s_bus = 20 m/s * t_bus

Setting these two expressions equal to each other:
20 m/s * t_bus = (1/2) * 0.8 m/s^2 * t_bus^2

Simplifying the equation:
0.8 * t_bus = 20

Solving for t_bus:
t_bus = 20 / 0.8
t_bus = 25 seconds

Therefore, it takes 25 seconds for the bus to catch up with the biker.