(A) 1.5 km Given, u = 0, v = 90 km/h = 90 x 5/18 = 25 m/s From the equation of motion, v = u + at a = v/t or, 25/2x60 or, 5/24 m/s² Distance covered is: S = ut + 1/2 at² = 0 + 1/2 x 5/24 x (120)² = 1/2 x 5/24 x 120 x 120 = 1500 m = 1.5 km

(A) 1.5 km
Given, u = 0, v = 90 km/h = 90 x 5/18 = 25 m/s
From the equation of motion,
v = u + at
a = v/t
or, 25/2×60
or, 5/24 m/s²
Distance covered is:
S = ut + 1/2 at²
= 0 + 1/2 x 5/24 x (120)²
= 1/2 x 5/24 x 120 x 120
= 1500 m
= 1.5 km

## Velocity time graph for a moving object is found to be curved line, then its acceleration is:

[B] variable

[B] variable

See less## A truck moving with a speed of 54 km/h. Truck driver applied brakes suddenly and brings the truck to rest in 5 s, then the average retarding force on truck, if mass of the truck and driver is 400 kg, will be:

(A) 1200 N

(A) 1200 N

See less## Intestine absorb the digested food materials. What type of epithelial cells are responsible for that?

(B) Columnar epithelium

(B) Columnar epithelium

See less## A Train starting from rest attains a velocity of 90 km/h in 2 min, then the distance travelled by the train for attaining this velocity is:

(A) 1.5 km Given, u = 0, v = 90 km/h = 90 x 5/18 = 25 m/s From the equation of motion, v = u + at a = v/t or, 25/2x60 or, 5/24 m/s² Distance covered is: S = ut + 1/2 at² = 0 + 1/2 x 5/24 x (120)² = 1/2 x 5/24 x 120 x 120 = 1500 m = 1.5 km

(A) 1.5 km

See lessGiven, u = 0, v = 90 km/h = 90 x 5/18 = 25 m/s

From the equation of motion,

v = u + at

a = v/t

or, 25/2×60

or, 5/24 m/s²

Distance covered is:

S = ut + 1/2 at²

= 0 + 1/2 x 5/24 x (120)²

= 1/2 x 5/24 x 120 x 120

= 1500 m

= 1.5 km

## A car retards uniformly at the rate of 5 m/s² and stops in 10 s. The initial velocity of car is:

(C) 50 m/s Given, velocity, v = 0, acceleration, a = - 5m/s² from equation of motion, v = u + at Or, u = - at = - (-5) × 10 = 50 m/s

(C) 50 m/s

See lessGiven, velocity, v = 0, acceleration, a = – 5m/s² from equation of motion, v = u + at

Or, u = – at = – (-5) × 10 = 50 m/s