Question

If acceleration-time graph of a particle which starts moving with initial velocity 10 m/s along positive X- axis is as given, then maximum speed of the particle is n m/s. Find the value of cms) (S) 2.​

Answer 1

Answer:

Solution

verified

Verified by Toppr

Correct option is

B

50

The distance travelled in n

th

second is

S

n

=u+

2

1

(2n−1)a ....(1)

So distance travelled in t

th

&(t+1)

th

second are

S

t

=u+

2

1

(2t−1)a ....(2)

S

t+1

=u+

2

1

(2t+1)a ....(3)

As per question,

S

t

+S

t+1

=100=2(u+at) ....(4)

Now from first equation of motion the velocity of particle after time t, if it moves with an acceleration a is

v=u+at ....(5)

where u is initial velocity

So from equation (4) and (5), we get v=50cm/s

Answer 2

Given info : If acceleration - time graph of a particle which starts moving with initial velocity, 10 m/s along positive x - axis is as given.

To find : the maximum speed of the particle is in m/s and also in cm/s ..

solution : first find equation of acceleration and time,

slope of graph = $\frac{y_2-y_1}{x_2-x_1}$ = $\frac{4-0}{0-2}$ = -2

now equation, (a - 4) = -2(t - 0)

⇒ a = 4 - 2t

⇒ dv/dt = 4 - 2t

⇒ ∫dv = ∫(4 - 2t)dt

⇒ v - u = 4t - t²

here initial velocity, u = 10 m/s.

∴ v = 4t - t² + 10

to get maximum value of v, dv/dt = 4 -2t = 0 ⇒ t = 2

and now, v = 4(2) - (2)² + 10 = 14 m/s

therefore the maximum velocity will be 14 m/s. and it is in cm/s is 1400 cm/s.