Question

A particle accelerates uniformly with a rate 10 m/s2 and then starts decelerating with a rate of 5 m/s2, if the total time of journey is 10 s, then the maximum velocity attained by it, is 10 3 m/s 30 m/s 3 100 m/s 3 40 m/s​

Answer 1

Solution

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Initial velocity u=30 km/h=30×

18

5

=

3

25

m/s

Final velocity v=60 km/h=60×

18

5

=

3

50

m/s

Mass m=1500 kg

Work done W=

2

1

mv

2

−

2

1

mu

2

⟹ W=

2

m

(v

2

−u

2

)=

2

1500

×((

3

50

)

2

−(

3

25

)

2

)=156187 J

Answer 2

Given: Initial rate of acceleration = 10 m/s²

           Rate of deceleration = 5 m/s²

           Total time of journey = 10 s

To Find : Maximum velocity attained by particle

Solution:

• For solving this question we will make use of the following formula:

                              v = u + at

Where,     v is final velocity of particle

               u is initial velocity of particle

               a is acceleration

        and t is time taken

• Here, we will make two equations:

1. for time $t_{1}$, when particle accelerates
2. for time $t_{2}$, when particle decelerates

• In the first equation, we assume the case that particle starts from rest and attains maximum velocity in time $t_{1}$.

                                  $v_{max}$ = 0 + 10$t_{1}$

                              ⇒ $v_{max}$ = 10$t_{1}$                           (1)

• In the second equation, we assume particle starts decelerating from its maximum velocity and comes to a stop in time $t_{2}$.

                              0 = $v_{max}$ - 5$t_{2}$                    

                               ⇒ $v_{max}$ = 5$t_{2}$                          (2)

              Equating (1) and (2) we get

                                10 $t_{1}$ = 5$t_{2}$

                             ⇒   2$t_{1}$= $t_{2}$                                       (3)

• We also know that total time = $t_{1}$ + $t_{2}$ = 10 s

                         Substituting values from (3)

                                    $t_{1}$ + 2$t_{1}$ = 10

                              ⇒ 3$t_{1}$ = 10

                              ⇒ $t_{1}$ = $\frac{10}{3}$ s                           (4)

• Substituting value of $t_{1}$ in (1) we get:

                       $v_{max}$ = 10 × $\frac{10}{3}$ = $\frac{100}{3}$ m/s

Therefore, maximum velocity attained by particle is $\frac{100}{3}$ m/s.