Question

A car is speeding at 25m/s in a low speed zone. A police car starts from rest just as car passes and accelerates at the constant rate of 5m/s^2. The police car will catch the speeding car after the time of (1)20s (2)10s (3)5s (4)15s

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Time\:taken=10\:sec}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

•. In the given question information given about a car is speeding at 25m/s in a low speed zone. A police car starts from rest just as car passes and accelerates at the constant rate of 5m/s^2.

• We have to find the time taken by police car to catch car.

$\underline \bold{Given : } \\ \bold{For \: car} \\ \implies Initial \: velocity(u) = 25 \: ms \\ \\ \bold{For \: police \: car : } \\ \implies Initial \: velocity = 0 \\ \\ \implies Acceleration(a) = 5 \: m {s}^{2} \\ \\ \underline \bold{To \: find: } \\ \implies Time \: taken \: to \: catch = ?$

• According to given question :

$\bold{For \: car : } \\ \implies s = ut + \frac{1}{2} a {t}^{2} \\ \\ \implies s = 25 \times t + \frac{1}{2} \times 0 \times {t}^{2} \\ \\ \implies s = 25t - - - - - (1) \\ \\ \bold{For \: police \: car : } \\ \implies s = ut + \frac{1}{2} {at}^{2} \\ \\ \implies s = 0 \times t + \frac{1}{2} \times 5 \times {t}^{2} \\ \\ \implies s = 2.5 \: {t}^{2} - - - - - (2) \\ \\ \bold{Comparing \: (1) \: and \: (2) \: we \: get : } \\ \implies 25 \: t = 2.5 \: {t}^{2} \\ \\ \implies \frac{t}{ {t}^{2} } = \frac{2.5}{25} \\ \\ \implies \frac{1}{t} = \frac{1}{10} \\ \\ \bold{\implies t = 10 \: sec} \\ \\ \bold{\therefore After \: 10 \: sec \: police \: jeep \: catch\: car}$

Answer 2

Solution:

Given:

➜ A car is speeding at 25m/s in a low speed zone. A police car starts from rest just as car passes and accelerates at the constant rate of 5m/s².

Find:

➜ Time taken by police car to catch the car.

Given:

For car:

➜ 25ms = initial velocity

For police car:

➜ 0 = initial velocity.

➜ 5 ms² = acceleation.

Know terms:

➜ Initial velocity = (v).

➜ Acceleation = (a).

According to given question:

For Car:

➜ s = ut + 1/2 at²

➜ s = 25 × t + 1/2 × 0 × t²

➜ s = 25t -----------------(1)

For Police car:

➜ s = ut + 1/2 at²

➜ s = × t + 1/2 × 5 × t²

➜ s = 2.5t² -----------------(2)

Comparing both equation (1) & (2), we get:

➜ 25t = 2.5t²

➜ t/t² = 2.5/25

➜ 1/t = 1/10

➜ t = 10

Therefore, 10 seconds is the time taken by police car to catch the car.

#AnswerWithQuality

#BAL