Question

A van is moving with the intial velocity of 'U' m/s after applying the brakes its retardation is 0.7 m/s2 and it stopped after 15 sec .find the intial velocity and distance travelled​

Answer 1

Answer:

Initial Velocity : 10.5 m/s

Distance Traveled : 78.75 m

Explanation :

Given :

1. $accelration = - 0.7{ms}^{ - 2}$
2. $time \: taken \: to\: stop \: = 15s$
3. $final\: veocity \: = 0 \: {ms}^{ - 1}$

To Find :

1. $initial \: velocty$
2. $distance \: travelled$

Solution :

We know that ,

$v \: = u \: + at$

$0 = u + ( - 0.7)(15)$

$u(initial \: velocity)= 10.5 \: {ms}^{ - 1}$

And ,

$s = ut \: + \frac{1}{2} {at}^{2}$

$s = 10.5(15) + \frac{1}{2} ( - 0.7) {15}^{2}$

$s = 157.5 - 78.75$

$s = 78.75 \: m$

Answer 2

Answer:-

• Distance travelled by van is 78.75m

__________

Explanation:-

Given that:

Initial velocity of car is u

Final velocity of car is 0m/s

[ As it is stopped by applying brakes ]

Time taken by car is 15s

Acceleration if van is -0.7m/s²

[ Retardation js negative acceleration ]

To Find: Distance travelled by van.

To calculate the distance travelled by van we must know that ::

First equation of motion: v = u - at

Third equation of motion: v² - u² = 2as

Where,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time taken
• s denotes distance covered

__________

~By applying first equation of motion.

➡ v = u + at

➡ 0 = u + ( -0.7 × 15 )

➡ 0 = u + ( -10.5 )

➡ u = 0 - (-10.5 )

➡ u = 10.5 m/s

~By applying third equation of motion.

➡ v² - u² = 2as

➡ ( 0 )² - ( 10.5 )² = ( -0.7 × 2 × s )

➡ 0 - 110.25 = -1.4 × s

➡ s = -110.25/-1.4

➡ s = 78.75m

__________

• Henceforth, the distance travelled by van is 78.75m