Question

A car starts from rest and accelerates uniformly to a speed of
180 kmh-1 in 10 seconds. The distance covered by the car in
this time interval is
[Kerala PMT 2009]
(a) 500 m
(c) 100 m
(b) 250 m
(d) 200 m
(e) 150 m​

Answer 1

Answer :

➥ The distance covered by a car = 250 m

Given :

➤ Initial velocity of a car = 0 km/hr (starts from rest)

➤ Final velocity of a car = 180 km/hr

➤ Time taken by a car = 10 sec

To Find :

➤ Distance covered by a car = ?

Required Solution :

✒ To find the Distance covered by a car, first we need to convert the Initial velocity and Final velocity of a bus km/hr to m/s, to convert the Initial velocity and Final velocity multiply the speed value by 5/18. after that we will find Acceleration of a bus, After all this we will find the Distance travelled by a bus.

◈ Initial velocity = 0 km/hr

→ Initial velocity = 0 × 5/18

→ Initial velocity = 0 m/s

◈ Final velocity = 180 km/hr

→ Final velocity = 180 × 5/18

→ Final velocity = 10 × 5

→ Final velocity = 50 m/s

❍ In the question, Initial velocity is given, Final velocity and Time taken is also given but we have to find the Distance covered by a car, so first we find Acceleration of a car.

This is given to find Acceleration of a car because of the first equation of motion:

★ v = u + at ★

So, we are provided with two elements used in the formula:

• Final velocity of a bus (v) = 50 m/s
• Initial velocity of a bus (u) = 0 m/s
• Time taken of a bus (t) = 10 sec

We have to find the Acceleration of a car.

From first equation of motion

⇒ v = u + at

⇒ 50 = 0 + a × 10

⇒ 50 = 0 + 10a

⇒ 50 = 10a

⇒ 50/10 = a

⇒ 5 = a

⇒ a = 5 m/s²

Now, we have three elements used in formula. Initial velocity, Time taken, and Acceleration of a bus.

Initial velocity of a bus = 10 m/s

Time taken by a bus = 10 sec

Acceleration of a bus = 1 m/s²

We can find Distance covered by a car by using the second equation of motion which says:

★ s = ut + ½ at² ★

Here,$\:$

• s is the Distance in m.
• u is the Initial velocity in m/s.
• t is the Time taken in second.
• a is the Acceleration in m/s².

✎ So, let's find Distance (s) !

From second equation of motion

⇛ s = ut + ½ at²

⇛ s = 0 × 10 + ½ × 5 × 10²

⇛ s = 0 + ½ × 5 × 10²

⇛ s = 0 + ½ × 5 × 100

⇛ s = 0 + ½ × 500

⇛ s = 0 + 1 × 250

⇛ s = 0 + 250

⇛ s = 250 m

║Hence, the distance covered by a car is 250 m.║

Answer 2

Question:-

A car starts from rest and accelerates uniformly to a speed of 180 kmh-1 in 10 seconds. The distance covered by the car in this time interval is [Kerala PMT 2009]

(a) 500 m

(b) 250 m

(c) 100 m

(d) 200 m

(e) 150 m

Answer:-

s = 250m

To Find:-

distance covered by the car in 10seconds.

Given:-

initial velocity (u) = 0

finally velocity (v) = 180 kmph –¹

time (t) = 10 seconds

acceleration (a) = ?

distance (s) = ?

★ First, we have to convert kmph to m/s.

So we have to multiply with 5/18

★ initial velocity (u) = 0 × 5/18

u = 0 [ something multiplied by zero is zero]

★ final velocity (v) = 180 × 5/18

v = 50m/s

Formulas:-

• v = u + at
• v² - u² = 2as
• s = ut + 1/2 at²

Solution:-

➭ v = u + at

➭ 50 = 0 + 10t

➭ 50 = 10t

➭ t = 50/10

➭ t = 5 seconds

Now, we have to find distance

➭ s = ut + 1/2 at²

➭ s = 1/2 × 5 × 10 × 10

➭ s = 250m

☞ Hence verified

Option B is the answer