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Grade 11Mechanics

A car accelerates from rest at a constant rate a for some time ,after which it decelerates at a constant rate b to come to rest .If total time elapsed is t second ,then calculate :
(i) the maximum velocity attained by the car
(ii)the total distance travelled by the car in terms of a,b and t.

Rakshith Krish

8 Years agoGrade 11

4 Answers

Arun

8 Years ago

Let the car accelerate for t₁sec at a m/sec² and decelerate for t₂sec. at - b/ sec²
At the end of t₁ sec. its velocity = 0 + at₁ m/sec. At the end of t₂ sec its velocity = 0
Initial velocity at the beginning of deceleration is α t₁
V(t₂) = a t₁ - b t₂ = 0 Therefore t₂ = (a/ b ) t₁ ------------------------(1)
But t = t₁ + t₂ ⇒ t = t₁ + t₂ = t₁ + (a/ b ) t₁ = t₁(1+ a/ b)
∴ t₁ = t / (1+ a/ b) = {b / (a + b)}*t -------------------------------------- (2)
Max. Velocity reached = a t₁ = a t / (1+ a/ b) = t / (1/ a + 1 / b)

= a*b*t/(a+b)
Distance covered =(a t₁² + b t₂²) /2 = ={a t₁² + b [(a/ b ) t₁)²]} /2 (substitute for t₂ from (1)
Simplifying this = ½ * (a/ b) * (a + b)* t₁²

Now substitute for t₁ from (2) to get
Distance covered = ½ *[a*b / (a +b)]*t²

Vimansh

7 Years ago

Total time equals T

a=v/t1

t1=v/a......1

t1 + t2 = T

t2=T-t1

B=u/T-t1

B=v/T-t1 (v in first case equals u in second case).....2

T=v/B+v/a

T=v(1/B + 1/a)

T=v(a+B/aB)

v=(aB/a+B)T

a is alpha and B is for beta

ankit singh

5 Years ago

For the first leg of the journey,
V = 0 + α t1
So V = α t1

for the second leg of journey of duration t - t1 :
0 = V - β (t - t1)
0 = α t1 - β t + β t1

t1 = β t / (α + β)

V = αt1 = α β t / (α + β )

Kinjal jaiswal

4 Years ago

Let max velocity be v

When a=α

v=u+at1

=0+αt1

Or T1= v/α

When a= β

V=u+at

=0+ βt2

Or T2= v/β

T1+T2=t= v/α+v/β

V= α βt/α +β

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