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tensa zangetsu [6.8K]
3 years ago
13

A woman on a motorcycle moving uniformly at a rate of 20 m/s passes a truck at rest. At the instant the motorcycle passes the tr

uck, the truck begins to accelerate at the rate of 5 m/s . (A) How long does it take the truck to catch up to the motorcycle? (B) How far has the motorcycle traveled? (C) What is the truck's speed at this point?
Physics
1 answer:
wlad13 [49]3 years ago
6 0

Answer:

A) It takes the truck 8 s to catch the motorcycle.

B) The motorcycle has traveled 160 m in that time.

C) The velocity of the truck is 40 m/s at that time.

Explanation:

The equations of the position and velocity of an object moving in a straight line are as follows:

x = x0 +v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

(A) When the the truck catches the motorcycle, both have the same position. Notice that the motorcycle moves at constant speed so that a = 0:

x truck = x motorcycle

x0 +v0 · t + 1/2 · a · t² = x0 + v · t

Placing the origin of the frame of reference at the point where the truck starts, both have an initial position of 0. The initial velocity of the truck is 0. Then:

1/2 · a · t² = v · t

solving for t:

t = 2 v/a

t = 2 · 20 m/s/ 5 m/s²

t = 8 s

It takes the truck 8 s to catch the motorcycle.

(B) Using the equation of the position of the motorcycle, we can calculate the traveled distance in 8 s.

x = v · t

x = 20 m/s · 8 s

x = 160 m

(C) Now, we use the velocity equation at time 8 s.

v = v0 + a · t

v = 0 m/s + 5 m/s² · 8 s

v = 40 m/s

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Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jaki
irina1246 [14]

1) See graph in attachment

2) 10 s

3) 50 m

Explanation:

1)

In this problem, we have an object initially moving with a velocity of

v = 10 m/s

when the time is

t = 0 s

Then, we are told that the speed of the object is decreasing by 1 m/s every  second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.

The result can be found in the graph in attachment.

Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to

a=1 m/s^2

which is equivalent to the gradient of the line in the velocity-time graph.

2)

In this part, we want to find after what time the body will stop its motion.

To do that, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

In this problem:

u = 10 m/s is the initial velocity of the body

a=-1 m/s^2 is the acceleration

v = 0 m/s, because we want to find the time T at which the body will stop

Re-arranging the equation, we find:

T=-\frac{u}{a}=-\frac{10}{-1}=10 s

3)

In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem:

u = 10 m/s is the initial velocity

a=-1 m/s^2 is the acceleration

t = 10 s is the time it takes for the body to stop (found in part 2)

Solving for s, we find the distance covered:

s=(10)(10)+\frac{1}{2}(-1)(10)^2=50 m

7 0
4 years ago
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