Question

A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β, to come to rest. If the total time elapsed is t, evaluate :- (a) the maximum velocity attained and (b) the total distance travelled. NO SPAM ✖✖

Answer 1

Solution:-The car will have max. velocity at the last moment of it's acceleration....

in first case

initial velocity=0

acceleration=a

let the time upto which it's accelerate be x

then,

final velocity during acceleration=u+at=ax

v=ax1

x1=v/a

now this final velocity will become ,initial velocity for the case of decelration

initial velocity=v

final velocity=0

time=x2 (say)

using first eq.

0=v-Bx2

x2=v/B

now as it is said that,total time taken is t,so

x1+x2=v/a +v/b

t=v(a+b)/ab

abt/(a+b)=v

ii) Distance travelled in the first case=a(x1)^2/2

=a×v^2/a^2×1/2=v^2/2a

distance travelled in second case=v^2/2a

=a^2b^2t^2/2a(a+b)

total distance travelled={v^2(a+b) +(abt)^2}/2a(a+b)

Answer 2

Question :-

A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β, to come to rest. If the total time elapsed is t, evaluate :-

(a) the maximum velocity attained and

(b) the total distance travelled.

Solution :-

(a) Let the car accelerates for time t₁ and decelerates for time t₂. Then,

t = t₁ + t₂          . . . (1)

(Check for the corresponding velocity–time graph in the attachment)

From the graph,

α = slope of line AB

$\sf{\implies \alpha=\dfrac{V_{max}}{t_1} }\\\\\sf{\implies t_1=\dfrac{V_{max}}{\alpha} }$

$\rule{120}1$

Again, from the graph,

β = -slope of line OB

$\sf{\implies \beta =\dfrac{V_{max}}{t_2} }\\\\\sf{\implies t_2=\dfrac{V_{max}}{\beta} }$

Putting the values of t₁ & t₂ in equation (1) :-

$\displaystyle{\sf{t=t_1+t_2}}\\\\\displaystyle{\sf{\hookrightarrow t=\frac{V_{max}}{\alpha } +\frac{V_{max}}{\beta} }}\\\\\displaystyle{\sf{\hookrightarrow t=V_{max}\bigg(\frac{1}{\alpha } +\frac{1}{\beta } \bigg)}}\\\\\displaystyle{\sf{\hookrightarrow t=V_{max}\bigg( \frac{\alpha + \beta }{\alpha \beta } \bigg)}}\\\\\boxed{\boxed{\sf{\color{coral}{\hookrightarrow V_{max}=\frac{\alpha \beta t}{\alpha+ \beta } }}}}$

$\rule{190}3$

(b) Total distance(s) = area under the V-T graph

$\sf{\hookrightarrow s=\dfrac{1}{2}\times V_{max} \times (t_1+t_2) }\\\\\sf{\hookrightarrow s= \dfrac{1}{2}\times \dfrac{\alpha \beta t}{\alpha + \beta } \times t }\\\\\boxed{\boxed{\sf{\color{coral}{\hookrightarrow s=\dfrac{1}{2}\bigg(\dfrac{\alpha \beta t^2}{\alpha + \beta } \bigg) }}}}$