Question

A car starts from rest and accelerates uniformly for 5seconds if it covers 200m in this duration , find its acceleration and final velocity

Answer 1

u = 0 m/s
t = 5 s
s = 200 m

v = ?
a = ?

Solution:-

s = ut + 1/2(at^2)
200 = 0(5) + 1/2(a)(25)
200 = 0 + 25/2(a)
200 = 25a/2
200 *2/25 = a
a = 8 ms^-2

now, 
u = 0 m/s
t = 5 s
s = 200 m
a= 8 m/s^2
v = ?

v = u + at
v = 0 + 8(5)
v = 40 m/s

Answer 2

Concept:

• One-dimensional motion
• Kinematics equations

Given:

• initial velocity u = 0 as the car starts from rest
• Duration of motion t = 5 s
• Distance covered in the motion s = 200 m

Find:

• The acceleration of the car, a
• The final velocity of the car, v

Solution:

We know the kinematics equation

s = ut +1/2at^2

s = 200 m, u = 0, t = 5 s

200 = 0 (5) + 1/2 a (5)^2

200 = 25/2 a

400 = 25 a

a = 400/25 = 16 m/s^2

Next, we use the following kinematics equation to determine the final velocity v

v = u+ at

u = 0, a = 16 m/s^2, t = 5 s

v = 0 + 16 (5)

v = 80 m/s

The acceleration is 16 m/s^2 and the final velocity is 80 m/s.

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