Question

A car travels with uniform velocity of 25 metre per second for 5 seconds. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 s. Find: (i) The distance which the car travels before the brakes are applied, (ii) Retardation and (iii) The distance travelled by the car after applying the brakes.​

Answer 1

Given :

A car travels with uniform velocity of 25 metre per second for 5 seconds. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 seconds.

To find :

• The distance which the car travels before the brakes are applied
• Retardation
• The distance travelled by the car after applying the brakes.

Solution :

• Initial velocity (u) = 25m/s
• Time (t) = 5s

As we know that,

⇒ Speed = distance/time

⇒ Distance = speed × time

⇒ Distance = 25 × 5

⇒ Distance = 125 m

Hence, distance travelled by car before applying brakes is 125 m.

_______________

• Initial velocity (u) = 25m/s
• Time (t) = 10s
• Final velocity (v) = 0 (brakes applied)

According to first equation of motion,

⇒ v = u + at

⇒ 0 = 25 + a × 5

⇒ 0 = 25 + 10a

⇒ 10a = - 25

⇒ a = -25/10 = -2.5m/s

Minus shows retardation of car. Hence, retardation of car is 2.5m/s.

_______________

• Initial velocity (u) = 25m/s
• Final velocity (v) = 0
• Acceleration (a) = - 2.5m/s

According to third equation of motion,

⇒ v² = u² + 2as

⇒ (0)² = (25)² + 2 × (-2.5) × s

⇒ 0 = 625 - 5s

⇒ 5s = 625

⇒ s = 625/5

⇒ s = 125 m

Distance travelled by the car after applying the brakes is 125m.

Answer 2

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Let us write the given components step by step

✧ It is given that Car travels with uniform velocity ✧

BEFORE APPLYING BRAKE:

✪ Initial Velocity (u) = 25 ms⁻¹

✪ Time (t) = 5 sec

AFTER APPLYING BRAKE:

✪ Final velocity (v) = 0 ms⁻¹

✪ Time (t) = 10 sec

TO FIND

(i) Retardation of Car

a = ?

✭ v = u + at

✭ v -u = at

✭ v - u/t = a

✭ $a = \frac{0-25}{10}$

✭ $a = \frac {-25}{10}$

✭ a = - 2.5

$\huge \fbox{ \underline \pink{ a = -2.5m/s }}$

Hence - 2.5 is the retardation of the car

(ii) The distance which the car travels before the brakes are applied

s = ?

Here accelration is unknown so we can use speed and distance formula

✧ Speed (U) = $\frac{Distance (s) }{Time (t)}$

✧ Speed × Time = Distance

✧ S = 25 × 5

✧ S = 125 meters

$\huge \fbox{ \underline \red{S = 125m }}$

✧ Hence car would have travelled 125 m before applying brakes

(iii) The distance travelled by the car after applying the brakes.

S = ?

Here we should take,

a = -2.5 m/s

t = 10 sec

u = 25 m/s

Using the Laws of Motion

✧ S = ut +$\frac{1}{2}$ at²

✧ S = 25×10 + $\frac{1}{2}$ × (-2.5) × 10²

✧ S = 250 + 50 × -2.5

✧ S = 250 - 125

✧ S = 125 meters

$\huge \fbox{ \underline \green{S = 125 m }}$

✧ Hence car would have travelled 125 m after applying brakes

$\underline{\underline{ \sf \huge \red{✪Extra Bytes✪}}}$

✭ Use the newtons 3 laws of motion accornding to given quantities and find the answers . They are,

✪ v = u + at ✪

✪ s = ut + 1/2 at²✪

✪ v² - u² = 2as ✪  

✭ With these formulas you can apply the given quantities and find the answers

✭ In some other cases Distance - Time formula also help in solving the problems

To learn more about this question refer

https://brainly.in/question/1159696

Hope it helps

Thanks for asking :-)