Question

Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jakim czasie ciało zatrzyma się. Oblicz przebytą drogę przez ciało podczas hamowania.( można skorzystać ze wzoru s=1/2at2gdzie a- opóźnienie, lub z wykresu)

Answer 1

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:

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