Question

E-1. The figure shows the velocity time graph of a particle which moves along a straight line starting withvelocity at 5 m/sec and coming to rest at t= 30s. Then :-A) Distance travelled by the particle is 212.5 m(B) Distance covered by the particle when it moves with constant velocity is 100 m (C) Velocity of the particle at t=25s is 5 m/sec(D) Velocity of the particle at t = 9s is 8 m/sec.question E-1. FAST PLEASE​

Answer 1

Concept:

The pace at which a body's distance changes in relation to time is referred to as its velocity.

Given:

The velocity-time graph of a particle that moves along a straight line with an initial velocity of $5m/s$ .

Find:

If the distance traveled by the particle is $212.5m$.

If the distance covered by the particle when moves with constant velocity is $100m$ .

If the velocity of the particle at $t=25s$ is $5m/s$ .

If the velocity of the particle at $t=9s$ is $8m/s$ .

Solution:

The total distance traveled by the particle is the difference between the area of the trapezium and the area of the triangle.

So,

The distance traveled $=\frac{1}{2}(30+60)(10)-\frac{1}{2}(5)(15) =212.5m$

Now,

The distance traveled by the particle with constant velocity is the area of the square:

$=5\times10=50m$

Now, at $t=25s$

The equation of the line CD is:

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

$v-v_1=\frac{v_2-v_1}{t_2-t_1}(t-t_1)$

$v-10=\frac{0-10}{30-20}(t-20)$

$v-10=-(25-20)\\v=5m/s$

Now,

At $t=9s$,

The equation of line OB is $y=mx+c$.

$v=mt+c\\v=\frac{y_2-y_1}{x_2-x_1}(9)+5$

$v=(9)\frac{10-5}{15-0}+5=8m/s$

Hence, the distance covered by the particle is $212.5m$, the velocity of the particle at $t=25s$ is $5m/s$ , and the velocity of the particle at $t=9s$ is $8m/s$.

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Answer 2

Answer:

Option A, C, D is correct and B is Wrong

Explanation:

A) Distance travelled by the particle is 212.5 m

 d₁ = 1/2 (5+10) (30)

     = 450/2

option A is correct

B) Distance covered by the particle when it moves with constant velocity is 100 m

d₂ = 10 x 5 = 50 m

option B is wrong

C) Velocity of the particle at t=25s is 5 m/sec

 a= 1 m/s²

V= U-at =10-5t

=5 m/sec

option c is correct

D) Velocity of the particle at t = 9s is 8

a=1/3  m/s²

v= 5+1/3 (9)

option D is correct

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