motion along a straight line problem...?

Question

the driver of a car wishes to pass a truck that is travelling at a constant speed of 20.0 m/s (about 45 mi/h). Initially, the car is also travelling at 20.0 m/s and its front bumper is 24.0m behind the truck's rear bumper. The car accelerates at a constant 0.600 m/s/s, then pulls back into the truck's lane when the rear of the car is 26.0 m ahead of the front of the truck. The car is 4.5m long and the truck is 21.0m long. A.) How much time is required for the car to pass the truck? B.) what distance does the car travel during this time? c.) what is the final speed of the car?

Answer 1

Assuming that Automan is driving the car (ie all lateral movements are at 90 degrees :)

Let us say that the maneuver takes t seconds.

The (front of the) car needs to travel 24m to catch up,, 21m to pass the truck and another 30.5 meters to cut in (=25+4.5)
This is = 75.5 meters travelled.

Since both vehicles are moving at initially the same speed, lets say they are stopped (relative to each other they are).

S = ut + 1/2 a t^2
75.5 = 0 + 1/2 * 0.6 * t^2
t^2 = 252
t = 15.86 seconds to pass the car

During this time in addition to the 75.5 meters the car (and truck) would travel the distance
S = ut + 0 = 20*15.86 = 317.28 m
Total distance travelled by car = 317.28+75.5 = 393m

The final speed of the car would be
v = u + 1/2 at
= 20 + .3*15.86
=25 m/s (approximately)

Answer 2

A) since u ask how much time it takes the car to pass the truck, then u must be talking abt the time period when the front of the car is just passing the back of the truck to the time when the back of the car is just passing the front of the truck....

so first let usfind the speed of the car when its front is aligned with the back of the truck....

since both vehicles have same initial speed, the relative velocity is zero..... (u=0)
acceleration, a = 0.6
displacement,s = 24m
we have v^2 = u^2 + 2as
by solving, we get v=5.367 m/s

now the relative velocity of the car with respect to the bus is 5.367m/s
the displacement (for passing the truck) equals the sum of the lengths of the vehicles.
s = 4.5 +21 = 25.5m
u = 5.367m/s
a = 0.6 m/s/s

we have s = ut + (1/2)at^2
by solving we can get the value of t, which is found out be 3.9 seconds
so the answer for the first part is t = 3.9 seconds

B) time , t = 3.9 s
initial velocity, u = 20+5.367 = 25.367m/s
acceleration, a = 0.6 m/s/s

we have s = ut + (1/2)at^2
by solving we can get the value of s
displacement, s = 103.4943 m

so the distance travelled by the car during the time of overtaking is 103.4943m

C) let us start from the beginning again.... the relative velocity at the beginning is taken as the initial velocity,u
u = 0 m/s
acceleration, a = 0.6 m/s/s
displacement can also be taken relative to the truck. it is the sum of the three distances given in the problem (1. the distance by which the car is behind, 2. the lengths of the car and truck together, 3.the distance by which the car is ahead)
therefore s = 24 + (4.5 + 21) + 26 = 75.5m

we have v^2=u^2 + 2as
by solving we can get the value of the final velocity, v as 9.518 m/s
this value of velocity is the relative velocity because the initial velocity and displacement are all given as relative to the truck.... sothe actual velocity of the car at the end will be given by (20+9.518)

final velocity of car = 29.518m/s


hope u r happy with my solution

Answer 3

a) it variety of feels we are going to ought to discover the area function first. We combine v(t) dt to discover the area x(t): x(t) = ? v(t) dt = ? (5t^2 - 10t) dt = (5/3)t^3 - 5t^2 at the same time as does x = 0? (5/3)t^3 - 5t^2 = 0 t^2((5/3)t-5)=0 t = 0, 3 So we verify the acceleration at t = 3. The acceleration function is given with suggestions from dv/dt: a(t) = dv/dt = 10t - 10 Plug in t = 3: a(3) = 20 m/s^2 b) Now we locate the minimum velocity. we are going to discover the intense factors of the speed function: 10t - 10 = 0 t = a million The minimum velocity takes position at t = a million and is: v(a million) = 5 - 10 = -5 m/s c) to discover this, we verify to carry close at the same time as v(t) is 0 back: 5t^2 - 10t = 0 5t(t - 2) = 0 t = 0, 2 So we verify to combine v(t) from t = a million to t = 2. This yields the following expression: (5/3)(2)^3 - 5(2)^2 - (5/3)(a million)^3 + 5(a million)^2 = 35/3 -15 = -10/3 m So the particle traveled 10/3 meters.