Question

A car initially at rest picks up a velocity of 72 kmh-1

in 1⁄4 minute. Calculate

(i) acceleration
(ii) distance covered by the car.

Answer 1

Answer:

According to the question let's first write down the given data,

Car at rest means the initial velocity of the car is given as, u= 0 km/hr

Then the final velocity, v = 72 km/hr after converting to m/s we have,

v = 72 × (5/18) = 20 m/s and the time taken,

t = 1/4 (60 sec) i.e. t = 15 sec

Now for us to calculate the distance covered by the car from above data, we should make use of one of the equations of motion,

To calculate the distance with constant acceleration and varying velocity we have the formula,

s = ut + 1/2 (at^2)

Now we have all the values except a,

To find a i.e. acceleration we have formula,

a = change in velocity/ time

i.e. a = (v-u)/t, a = (20–0)/15 = 1.33 m/s

Now, substituting the values in above equation

We get, s = (0) × (15) + 1/2 (1.33) × (15)^2

s = 150m

So, the distance covered by the car is 150m.

Answer 2

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Speed = 72 km/hr = 72×(5/18) = 20 m/s

Acceleration a is obtained from the formula " v = u + a×t " ,

with initial speed u = 0 and final speed v = 20 m/s. t is time taken, 20 s.

a = v/t = 20/20 = 1 m/s2

Force = mass×acceleration = 1000×1 = 1000 N

distance travelled S is obtained from formula " S = u×t+(1/2)×a×t2 " , with u =0, t =20 s

S = (1/2)×1×20×20 = 200 m