Question

a car starts from rest and accelerates from rest uniformly for 10s to a velocity of 36km/hr. it then runs at a constant velocity and is finally brought to rest in 50m with uniform retardation.if the total distance covered by the car is 500m. find its retardation​

Answer 1

Answer:

Refer to Image)

V=at

⇒8=a10

⇒a=0.8m/s

2

V

2

=U

2

+2as

⇒U=8

2

+2×a64

⇒a=

2×64

−64

=−0.5m/s

2

Distance travelled= Area of graph

⇒584=

2

1

×10×8+8×t

1

+

2

1

×16×8

⇒584=40+8t

1

+64

⇒8t

1

=480

⇒t

1

=60

Total time= 10+60+16=86s

Answer 2

$Given,$

$Initial \: Velocity \: u = 0 \: m \: {s}^{ - 1}$

$Final \: Velocity \: v = 36 \: km \: {h}^{ - 1} \\ \: \: \: \: \: \: \: \: \: \: \: = \frac{36 \times 1000}{3600} \frac{m}{s} = 10 \: m \: {s}^{ - 1}$

$Time \: t = 10 \: s$

$Acceleration \: a = \frac{v - u}{t}$

$Acceleration \: a = \frac{10 - 0 \: m \: {s}^{ - 1} }{10 \: s}$

$Acceleration \: a = 1 \: m \: {s}^{ - 2}$

$Now,$

$Initial \: Velocity \: u = 10 \: m {s}^{ - 1}$

$(because \: the \: car \: runs \: at \: a \: constant \: velocity \: u \: )$

$Final \: Velocity \: v = 0 \: m \: {s}^{ - 1}$

$Distance \: S = 50 \: m$

$According \: To \: Relation \: ...$

${v}^{2} = {u}^{2} + 2aS$

$0 = {10}^{2} m \: {s}^{ - 0} - 2a \times 50m$

$a = \frac{ - 100}{2 \times 50} = - 1 \: m \: {s}^{ - 2}$

$Therefore \: Retardation = 1 \: m \: {s}^{ - 2}$