Question

A car travels with uniform velocity of 20m/s for 5s. The brakes are then applied and the car is uniformly retarded. It comes to rest in further 8s. Find -
a) Distance traveled in first 5s
b) Distance traveled after brakes are applied
c) Total distance traveled
d) Acceleration during the first 5s and last 8s

Answer 1

Initial Velocity = 20 m/sec

Time = 5 seconds.

After 5 seconds

Initial velocity = 20 m/sec

Final velocity = 0m/sec

Time = 8seconds

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(i) Distance covered in first 5 seconds

Since Body moved with uniform motion, this means that acceleration =0

So,From 2nd equation of motion :-

$s = ut + \frac{1}{2}a {t}^{2} \\ \\ \\s = ut \: \: \: \: (as \: \: a = 0) \\ \\ \\s = 20 \times 5 \\ \\ \\s = 100m$

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(ii) Distance covered after breaks are applied :-

U= 20 m/sec

V = 0 m/sec

T = 8 second

So,

$v = u + at \\ \\a = \frac{v - u}{t} \\ \\a = \frac{ - 20}{8} = \frac{ - 5}{2} = - 2.5$

So,

$s = \frac{ {v}^{2} - {u}^{2} }{2a} \\ \\ \\ \\s = \frac{ {- (20)}^{2} }{ - 5} \\ \\ \\s = \frac{400}{5} \\ \\ \\s = 80m$

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(iii) Total Distance travelled

$= 100 + 80 \\ \\ \\ = 180m$

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(iv)

Acceleration during 1st 5 seconds = 0m/sec²

Acceleration during last 8 seconds = (-2.5)m/sec²

Answer 2

Motion in Straight Line :

Initial velocity, u = 20 m/s

Time taken, t = 5s

Time taken by car to retãrd uniformly, t2 = 8s

( a ). Distance travelled in first 5 seconds

Using 2nd Equation of motion :

$\boxed{\mathsf{\green{s\:=\:ut\:+\:{\dfrac{1}{2}*a{t}^{2}}}}}$

$\mathsf{s\:=\:20*5\:+\:{\dfrac{1}{2}*0*{5}^{2}}}$

[ Acceleration, a = 0, because car is uniformly retãrded ]

$\boxed{\mathsf{s_1\:=\:100\:m}}$

( b ). Distance travelled after applying the brakes,

First we will find acceleration / Retardation,

Using 1st Equation of motion ,

$\boxed{\mathsf{\green{v\:=\:u\:+\:at}}}$

$\mathsf{0\:=\:20\:+\:a*8}$

$\mathsf{a\:=\:{\dfrac{-20}{8}}}$

$\mathsf{a\:=\:-2.5\:m{s}^{-2}}$

Now, Using 3rd Equation of motion,

$\boxed{\mathsf{\green{v^2\:-\:u^2\:=\:2as}}}$

$\mathsf{0\:-\:20^2\:=\:2*(-2.5)s}$

$\mathsf{s\:=\:{\dfrac{400}{5}}}$

$\boxed{\mathsf{s_2\:=\:80\:m}}$

( c ). Total distance travelled = $\mathsf{s_1\:+\:s_2}$

Total distance travelled = 100 + 80 = 180 m.

( d ). Acceleration during the first 5s and last 8s = $\mathsf{- 2.5 \:m{s} ^{-2}}$ [ From part ( b )].