Question

A cheeta starts from rest,and acceleration 2m/s² for 10 seconds.Calculate—:
(a)the final velocity
(b)the distance travelled

Answer 1

Initial velocity = u = 0m/s
final velocity = v= ?
acceleration = a = 2m/s²
time = t = 10 s
distance = s = ?

(a)
a = (v-u)/t
at = v-u
at + u = v
2×10 + 0 = v
20 m/s = v

final velocity of the cheeta is  20m/s

(b)
v² - u² = 2as
(20)² - (0)² = 2(2)s
400 = 4s
400/4 = s
100 m = s

the distance traveled by the cheeta is 100 metre
 

Answer 2

Given,

Initial velocity, u = 0
Acceleration, a = 2m/s²
Time, t = 10s

Now,

From 1st equation of motion,

v = u + at

Or v = 0 + 2×10

Or v = 20 m/s

Again,
From 3rd equation of motion,

v² - u² = 2as

Or (20)² - (0)² = 2×2s

Or 400 = 4s

Or s = 100 m

Thus, (a) the final velocity is 20 m/s and (b) the distance travelled is 100m