Question

A car moves in a straight line with a constant acceleration of 2 m/s². It changes its velocity from 10m/s to 20m/s in t1 seconds. Then after it accelerates with an acceleration of 5m/s² and its velocity reduces to 10m/s in t2 second again. Find the value of t1 × t2.​

Answer 1

Answer:

10sec

Explanation:

initial velocity =10m/s

after first acceleration of 2m/s² final velocity is 20m/s in time t1.

as it constant Acceleration case,

dv/dt = ∆v/∆t = a

= (20-10)/t1 = 2

=> t1 = 5sec

after that it is retarded at 5m/s² to velocity of 10m/s in time t2

so, (10-20)/t2 = -5 {- because of Retardation}

=> t2 = 2 sec

hence t1 x t2 = 10sec

You can also use first equation of motion as it is case of uniform acceleration case.

Hope this will help

Answer 2

${\large{\underline{\pmb{\sf{Given...}}}}}$

★ A car moves in a straight line with a constant acceleration of 2 m/s²

★ It changes its velocity from 10m/s to 20 m/s in t$_1$ seconds

★ Then after it accelerates with an acceleration of 5m/s² and its velocity reduces to 10m/s in t$_2$ seconds

${\large{\underline{\pmb{\sf{To \; Find...}}}}}$

★ The value of t$_1$ × t$_2$ as per the above given statements

${\large{\underline{\pmb{\sf{Understanding \; the \; Concept...}}}}}$

❍ Concept : Here we have been given the initial,the final velocity and the acceleration in the first case which are 10m/s, 20m/s and 2 m/s² respectively similarly in the second case we have been provided with the acceleration, initial velocity and the final velocity  in the second case which are 5m/s , 20m/s and 10m/s  respectively

⋆ Now as we have to find out the value of t$_1$ × t$_2$ so, let's use a suitable formula to find the the time in the first and the second case so that later we can multiply the time taken by the car and find out the required answer as per the need of thee question

${\large{\underline{\pmb{\sf{Using \; Concept...}}}}}$

✪ Formula to find out acceleration :

$\tt Acceleration = \dfrac{v-u}{t}$

${\large{\underline{\pmb{\sf{Solution...}}}}}$

★ The value of t$_1$ × t$_2$ is 10 seconds²

${\large{\underline{\pmb{\sf{Full \; Solution...}}}}}$

~ In the first case as we have the initial velocity, final velocity and the acceleration so, let's find out the time using the above mentioned formula

→ Formula,

• a = v - u / t

→ Where,

• a = Acceleration
• v = Final velocity
• u = Initial velocity
• t = Time

~ Now let's substitute the values in the formula and find t$_1$

${:\implies}\sf a = \dfrac{v-u}{t}$

${:\implies}\sf 2 = \dfrac{20 - 10}{t_1}$

${:\implies}\sf 2 = \dfrac{10}{t_1}$

${:\implies}\sf t_1 = \dfrac{10}{2}$

${:\implies}\sf t_1 = 5 \; seconds$

• Henceforth the time taken is 5 seconds

~ Now let's find the time taken in the second case

${:\implies}\sf a = \dfrac{v-u}{t}$

${:\implies}\sf - 5 = \dfrac{10 - 20}{t_2}$

${:\implies}\sf -5 = \dfrac{-10}{t_2}$

${:\implies}\sf t_2 = \dfrac{-10}{-5}$

${:\implies}\sf t_2 = 2 \; seconds$

• Henceforth the time taken is 2 seconds

~ Now let's multiply t$_1$ × t$_2$ and find the required answer

${:\implies}\sf t_1 \times t_2$

${:\implies}\sf 5 \; s \times 2\;s$

${:\implies}\sf 10\; seconds \; ^2$

• Henceforth the value of t $_1$ × t $_2$ is 10 seconds²