Question

Find the velocity of a car which starts from rest and moves with constant acceleration 10 ms-2 for 1.5s. What distance will the car cover in 1.5s?​

Answer 1

Answer:

Suppose car acceleration for time t. So it decelerate by time (15−t)s.

Initially at rest, u=0 and come to velocity v in time t! use from Kinematics

for constant accim, (a=10m/m² )

⇒ v=⊥0t →(1)

⊥ & (2) gives,

⊥Ot=5(15−t) ⇒ t=5 s.

for constant deceleration (a=−5m/s)

⇒ o=v−5(15−t)

⇒ v=5(15−t) →(2)

so, v=⊥ot=50 m/s

construct a v−t graph from obtained results.

From graph

(i) max velocity =50 m/s

(ii) net distance travelled =area ot vt graph

d= ½×50×15=375 m

∴ average velocity = t÷d= 375÷15 =25 m/s

during deceleration distance travelled,

d1= ½ ×(15−5)×50=250m

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Answer 2

Given :      a car which starts from rest and moves with constant acceleration 10 ms-2 for 1.5s.

To Find : velocity after 1.5 sec

distance will the car cover in 1.5s?

Solution:

v = u + at

u = 0 m/s

a = 10 m/s²

t = 1.5 sec

v = 0 + 10 (1.5)

=> v = 15

Hence velocity after 1.5 sec  = 15 m/s

S = (1/2)(u + v) * t

=> S = (1/2)(0 + 15) * 1.5

=> S = 11.25  m

or use

S = ut + (1/2)at²

=> S = 0 + (1/2) (10) (1.5)²  =   11.25  m

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