Question

A car moves at a uniform acceleration along a straight lines PQR . Its speed at P and R are 5 m/s and 25m/s respectevely. If pq:qr =1:2, the ratio of time taken by the car to travel distance pq and qr is :
a) 1:2
b)2:1
c)1:3
d)1:1​

Answer 1

answer : option (4) 1 : 1

explanation : let uniform acceleration is a m/s² and velocity at Q is v m/s.

velocity at P = 5m/s

velocity at R = 25 m/s

using formula, v² = u² + 2as

for path PQ : final velocity = v , initial velocity = 5m/s and PQ = s

so, v² = (5)² + 2as........(1)

for path QR : final velocity = 25m/s, initial velocity = v and QR = 2PQ = 2s

so, (25)² = v² + 2a(2s)

⇒ 625 = v² + 4as ........(2)

from equations (1) and (2),

625 - 2v² = v² - 2(5)²

⇒625 + 50 = 3v²

⇒3v² = 675

⇒v² = 225

⇒v = 15 m/s

and 2as = (15)² - (5)² ⇒a = 100/s [ from eq (1) ]

now time taken to travel distance PQ , t = (v - 5)/a

= (15 - 5)/(100/s) = 10s/100 = s/10....(3)

time taken to travel distance QR , t = (25 - v)/a = (25 - 15)/(100/s) = s/10 .....(4)

from equations (3) and (4),

ratio of time taken by the car to travel distance PQ and QR is s/10 : s/10 = 1 : 1

Answer 2

Answer:

Option (d) 1:1

Explanation:

Initial speed of the car at P

$u=5$ m/s

Speed of the car at R (Final speed)

$v=25 m/s$

If the common ratio is x then

Distance $PQ=x$

Distance $QR=2x$

Therefore, total distance

$s=x+2x=3x$

Let the constant acceleration be a

From the third equation of motion

$v^2=u^2+2as$

$25^2=5^2+2a\times 3x$

$\implies ax=100$

If the velocity at Q is v' then

$v'^2=u^2+2a(x)$

$\implies v'^2=5^2+2\times 100$

$\implies v'^2=25+200$

$\implies v'=225$

$\implies v'=15$ m/s

If time taken in travelling from P to R is T then

From the first equation of motion

$v=u+aT$

$25=5+aT$

$\implies T=20/a$

If the time taken in travelling from P to Q is $t_1$ then

$v'=u+at_1$

$\implies 15=5+at_1$

$\implies at_1=10$

$\implies t_1=10/a$

Therefore, time taken in travelling from Q to R

$t_2=T-t_1=20/1-10/a=10/a$

Ratio

$\frac{t_1}{t_2}=1$

or, $t_1:t_2=1:1$

Hope this helps.