Question

6 m/s2 acceleration is given to a body moving with a velocity 30 m/s 4x4=16
What is the time needed for it to come to rest?

Answer 1

Answer:

We will start with calculating the final velocity:

Vf = Vi + a×t

Vf= Final Velocity (m/s)

Vi = Initial Velocity (m/s)

a = Acceleration (m/s^2)

t = Time (seconds)

We will plug in the appropriate information into the formula:

Vf = (6 m/s) + ((1.5 m/s^2)×(5 seconds))

Vf= (6 m/s) + (7.5)

Vf= (13.5 m/s)

We will calculate the distance covered:

D= (Vi)×(t) + (1/2)×(a)×(t^2)

D= distance (meters)

vi= initial velocity (m/s)

t=time (seconds)

a= acceleration (m/s/s or m/s^2)

D = (6 m/s)×(5 sec) + (1/2)×(1.5 m/s^2)×(5^2)

D= (30)+ (18.75)

D= (48.75 meters)

I hope this helps! All the best!

Answer 2

Given:

A body moving with 30 m/s velocity gets acceleration of -6 m/s².

To find:

Time to stop ?

Calculation:

• Let the time to stop be 't':

Applying Equation of Kinematics:

$\rm \: v = u + at$

$\rm \implies \: 0 = 30 + ( - 6)t$

$\rm \implies \: 30 - 6t = 0$

$\rm \implies \: 6t = 30$

$\rm \implies \: t = \dfrac{30}{6}$

$\rm \implies \: t = 5 \: sec$

So, time taken to stop is 5 seconds.