A car accelerates uniformly from rest to a speed of 36 km/hr in 5 seconds. The distance covered by the car in the first 10 seconds is:
Answers
Answer:
Given parameters
Time taken (t) = 5 sec
Initial velocity (u) =18 km/hour
\(u = \frac{18\times 1000}{60\times 60s}\)
u = 5 m/s
Final velocity (v) =36km/hour
\(v = \frac{36\times 1000}{60\times 60s}\)
v =10 m/s
We need to calculate acceleration and distance traveled
We know that
acceleration a = (v – u)/t
Substituting the given values in the above equation we get,
a = (10 – 5)/5
a =5/5
a = 1 m/s2
We know that distance travelled is calculated by the formula
Distance travelled S = u t + (1/2) a × t2
Substituting the given values in the above equation we get,
S = 5 × 5 + (1/2) × 1 × 52
S = 25 + (1/2) × 25
S = 25 + 12.5
S = 37.5 m
Hence
acceleration a =1 m/s2
Distance travelled S =37.5 m
Answer:
A car accelerates uniformly from 18km per hour to 36km per hour in 5 seconds . Calculate the acceleration and the distance covered by the car in that time.
Answer:
Given parameters
Time taken (t) = 5 sec
Initial velocity (u) =18 km/hour
\(u = \frac{18\times 1000}{60\times 60s}\)
u = 5 m/s
Final velocity (v) =36km/hour
\(v = \frac{36\times 1000}{60\times 60s}\)
v =10 m/s
We need to calculate acceleration and distance traveled We know that
acceleration a = (v – u)/t
Substituting the given values in the above equation we get,
a = (10 – 5)/5
a =5/5
a = 1 m/s2
We know that distance travelled is calculated by the formula
Distance travelled S = u t + (1/2) a × t2
Substituting the given values in the above equation we get,
S = 5 × 5 + (1/2) × 1 × 52
S = 25 + (1/2) × 25
S = 25 + 12.5
S = 37.5 m
Hence
acceleration a =1 m/s2
Distance travelled S =37.5 m