Question

A car moves with uniform
velocity of 40 ms for 5 s. It
Comes to rest in the next 10s with uniform deceleration. Find(i) deceleration (ii) total
distance travelled by the car.


With detail
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Answer 1

Answer:

Given data:

• Initial velocity = Vi = 40m/s
• Final velocity = Vf = 0m/s
• Time. = t1. = 5 sec
• t2 = 10 sec

To find:

• deceleration = d = ?
• Total distance = S = ?

Solution:

First we have to find the declaration.

deceleration is the negative of acceleration.

By using first equation of motion

$\large{Vf = Vi + at}$

8Put values in the formula,

$\large{ \mathscr{0 = 40 + a(10)}} \\ \large{ \mathscr{ - 40 = a(10)}} \\ \large{ \mathscr{ \frac{ - 40}{10} = a}} \\ \large{ \mathscr{a = - 4m {s}^{ - 2} }}$

Now by using third equation of motion.

${2aS = {Vf}^{2} - {Vi}^{2}}{{}}$

put values

$2( - 4)S = {0}^{2} - {40}^{2} \\ - 8S = 0 - 1600 \\ - 8S = - 1600 \\S = \frac{ - 1600}{ - 8} \\ S _{1} = 200m$

It was S1 not the total distance if we have to find the total distance we have to use the formula of distance.

$S = Vi × t</p><p> \\ \: \: \: \: \: \: = 40 \times 5 \\ S _{2} = 200m$

Total distance can be found by multiplying these both distances.

$\mathscr{total \: distance} =S_{1}+ S_{2}</p><p> \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =200 + 200 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 400m$

So,

Deceleration = -4 m/s²

total distance = 400m