Question

4.
A jeep starts from the state of rest. If its velocity becomes 60 km/hr in 5 minutes,
(i) What is the acceleration of the jeep? (ii) What is the distance covered by the
jeep?
​

Answer 1

Answer:

$\boxed{\sf Acceleration \ of \ jeep = 200 \ m/min}$

$\boxed{\sf Distance \ covered \ by \ jeep = 2500 \ m \ \ or \ \ 2.5 \ km}$

Given:

Initial velocity (u) = 0 km/hr

Final velocity (v) = 60 km/hr

Time taken (t) = 5 min

To find:

(i) Acceleration of the jeep

(ii) Distance covered by jeep

Explanation:

Let's first convert velocity from km/hr to m/min;

$\sf \implies 60 \ km/h = \cancel{60} \times \frac{1000}{ \cancel{60}} \ m/min \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1000\ m/min$

So,

Final velocity (v) = $\sf \frac{50}{3} \ m/min$

$\sf (i) From \ 1^{st} \ equation \ of \ motion: \\ \sf \implies v = u + at \\ \\ \sf \implies 1000 = 0 + a(5) \\ \\ \sf \implies a \times 5 = 1000 \\ \\ \sf \implies a = \frac{1000}{5} \\ \\ \sf \implies a = \frac{200 \times \cancel{5}}{\cancel{5}} \\ \\ \sf \implies a = 200 \ m/min^{2}$

$\sf (ii) From \ 2^{nd} \ equation \ of \ motion: \\ \sf \implies s = ut + \frac{1}{2} at^{2} \\ \\ \sf \implies s = 0(5) + \frac{1}{2} \times200\times {5}^{2}\\ \\ \sf \implies s = 0 + \frac{1}{2} \times 200 \times 25 \\ \\ \sf \implies s = \frac{1}{ \cancel{2}} \times \cancel{2} \times 100 \times 25 \\ \\ \sf \implies s = 100 \times 25 \\ \\ \sf \implies s = 2500 \: m\\ \\ \sf \implies s = \frac{2500}{1000} \:km\\ \\ \sf \implies s = 2.5 \: km$

Answer 2

Given:-

• Initial velocity of jeep (u) = 0

• Final velocity of jeep (v) = 60km/h = 60×5/18=50/3 m/s

• Time taken (t) = 5 min. = 5×60 = 300s

To Find:-

(i) The acceleration (a) of the jeep.

(ii) Distance (s) covered by jeep.

Solution :-

(i)

By using 1s equation of motion

➦ v = u +at

➭ 50/3= 0+ a× 300

➭ a = 50/3×300

➭ a = 5/90m/s²

∴ The acceleration of jeep is 5/90 m/s²

_______________________________

(ii)

By using 2nd equation of motion

➦ s = ut+1/2at²

➭ s = 0×300+1/2×5/90× 300×300

➭ s = 1/2×5000

➭ s = 2500m

☛ 2500/1000= 2.5km

∴ The distance covered by jeep is 2.5km