Question

find out distance travelled by the block in 10s for the given situation ​

Answer 1

Given, initial velocity (u) = 10m/s

acceleration (a) = - 2m/s²

To find : Distance traveled in 10 seconds.

Finding final velocity for the given time.

1st Equation of motion

$v = u + at$

Final velocity after 10 seconds

v = 10 + (-2)(10)

v = - 10 m/s

This shows the body is going to move in the opposite direction. So Distance ≠ Displacement and we can't use the second equation of motion directly.

From first equation of motion, Velocity after t seconds can be given by,

v = 10 - 2t

Finding the time at which velocity becomes 0 ( If velocity becomes zero for a body on a straight path, we may consider it to be changing direction)

So,

0 = 10 - 2t

10 = 2t

t = 5 seconds.

Therefore, The body travels for 5 seconds in the current direction, later it changes it's direction of motion.

Distance traveled in 10 seconds

= Distance traveled in the first 5 seconds + Distance traveled in the second 5 seconds.

For the first 5 seconds

u = 10m/s

a = - 2m/s²

t = 5 s

S = ut + ½at²

S = 10(5) + 1/2 ( - 2)(5²)

S = 50 - 25

S = 25 m.

For the last 5 seconds,

u = 0 m/s

a = - 2 m/s²

t = 5s

S = ut + 1/2at²

S = 0(5) + 1/2 ( - 2)(5)²

S = - 25m.

The body travels this distance in the opposite direction of initial motion.

Therefore, The total distance traveled by the body in 10 seconds is 25 + 25 = 50m.

Answer 2

Answer:

Given:

Initial Velocity of the object = 10 m/s

Acceleration = - 2 m/s² ( negative sign means opposite direction as that of initial Velocity )

To find :

Distance travelled in 10 seconds.

Concept:

First of all , Distance is a scalar Quantity. It has only magnitude, no Direction. Hence it can never be zero ( always positive).

If we try to draw a speed vs the graph, we can say that the area under the speed time graph shall give us Distance.

Mathematically , it can be denoted as :

$\boxed { \huge{ \red{x= \displaystyle \int \: v \: dt}}}$

Calculation:

Let's try to frame an equation :

Let final Velocity be v

$\boxed{ \huge{ \green{v = u + at}}}$

$= > v = 10 + ( - 2)t$

$= > v = 10 - 2t \: .....(1)$

Comparing this equation with

$\boxed{ \huge{ \blue{ \bold{y = mx + c}}}}$

We have positive intercept and negative slope.

So we get the graph as attached in the photo.

• Put t = 0 s , we get v = 10 m/s

• Put v = 0 m/s , we get t= 5 s

Now after t = 5 seconds, the Velocity time graph goes below the + y axis.

Put t = 10 s , we get v = -10 m/s

So distance is given by the total area under the Velocity time graph :

$d = sum \: of \: area \: of \: 2 \: \Delta s$

$= > d = \bigg \{\frac{1}{2} \times 10 \times 5 \bigg \} + \bigg \{ \frac{1}{2} \times 10 \times 5 \bigg \}$

$= > d = 25 + 25$

$= > d = 50 \: metres$

So final answer :

$\boxed{ \sf{ \orange{ \huge{d = 50 \: m}}}}$